how to solve a probability question

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Mathematics
Probability and Statistics

How to Calculate Probability

Last Updated: October 1, 2022 References Approved

This article was co-authored by Mario Banuelos, PhD . Mario Banuelos is an Assistant Professor of Mathematics at California State University, Fresno. With over eight years of teaching experience, Mario specializes in mathematical biology, optimization, statistical models for genome evolution, and data science. Mario holds a BA in Mathematics from California State University, Fresno, and a Ph.D. in Applied Mathematics from the University of California, Merced. Mario has taught at both the high school and collegiate levels. There are 8 references cited in this article, which can be found at the bottom of the page. wikiHow marks an article as reader-approved once it receives enough positive feedback. This article has 14 testimonials from our readers, earning it our reader-approved status. This article has been viewed 3,302,977 times.

Chances are (pun intended) you've encountered probability by now, but what exactly is probability, and how do you calculate it? Probability is the likelihood of a specific event happening, like winning the lottery or rolling a 6 on a die. Finding probability is easy using the probability formula (the number of favorable outcomes divided by the total number of outcomes). In this article, we'll walk you through exactly how to use the probability formula step by step, plus show you some examples of the probability formula in action.

Finding the Probability of a Single Random Event

Image titled Calculate Probability Step 1

Example: It would be impossible to calculate the probability of an event phrased as: “Both a 5 and a 6 will come up on a single roll of a die.”

Image titled Calculate Probability Step 2

Image titled Calculate Probability Step 3

Image titled Calculate Probability Step 4

For example, the likelihood of rolling a 3 on a 6-sided die is 1/6. But the probability of rolling all five other numbers on a die is also 1/6. 1/6 + 1/6 + 1/6 + 1/6 + 1/6 + 1/6 = 6/6 , which = 100%.

Note: If you had, for example, forgotten about the number 4 on the dice, adding up the probabilities would only reach 5/6 or 83%, indicating a problem.

Image titled Calculate Probability Step 5

For example, if you were to calculate the probability of the Easter holiday falling on a Monday in the year 2020, the probability would be 0 because Easter is always on a Sunday.

Calculating the Probability of Multiple Random Events

Image titled Calculate Probability Step 6

Note: The probability of the 5s being rolled are called independent events, because what you roll the first time does not affect what happens the second time.

Image titled Calculate Probability Step 7

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Converting Odds to Probabilities

Image titled Calculate Probability Step 9

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The event that you’ll draw a white marble is 11; the event another color will be drawn is 9. The total number of outcomes is 11 + 9, or 20.

Image titled Calculate Probability Step 11

So, in our example, the probability of drawing a white marble is 11/20. Divide this out: 11 ÷ 20 = 0.55 or 55%.

Probability Cheat Sheets

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About This Article

Probability is the likelihood that a specific event will occur. To calculate probability, first define the number of possible outcomes that can occur. For example, if someone asks, “What is the probability of choosing a day that falls on the weekend when randomly picking a day of the week,” the number of possible outcomes when choosing a random day of the week is 7, since there are 7 days of the week. Now define the number of events. In this example, the number of events is 2 since 2 days out of the week fall on the weekend. Finally, divide the number of events by the number of outcomes to get the probability. In our example, we would divide 2, the number of events, by 7, the number of outcomes, and get 2/7, or 0.28. You could also express the answer as a percentage, or 28.5%. Therefore, there’s a 28.5% probability that you would choose a day that falls on the weekend when randomly picking a day of the week. To learn how to calculate the probability of multiple events happening in a row, keep reading! Did this summary help you? Yes No

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How to Solve Probability Problems? (+FREE Worksheet!)

Do you want to know how to solve Probability Problems? Here you learn how to solve probability word problems.

Step by step guide to solve Probability Problems

Probability Problems – Example 1:

If there are $8$ red balls and $12$ blue balls in a basket, what is the probability that John will pick out a red ball from the basket?

There are $8$ red balls and $20$ a total number of balls. Therefore, the probability that John will pick out a red ball from the basket is $8$ out of $20$ or $\frac{8}{8+12}=\frac{8}{20}=\frac{2}{5}$.

Probability Problems – Example 2:

A bag contains $18$ balls: two green, five black, eight blue, a brown, a red, and one white. If $17$ balls are removed from the bag at random, what is the probability that a brown ball has been removed?

If $17$ balls are removed from the bag at random, there will be one ball in the bag. The probability of choosing a brown ball is $1$ out of $18$. Therefore, the probability of not choosing a brown ball is $17$ out of $18$ and the probability of having not a brown ball after removing $17$ balls is the same.

Exercises for Solving Probability Problems

Download Probability Problems Worksheet

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Tricks To Solve Probability Questions

The application or uses of probability can be seen in quantitative aptitude as well as in daily life. It is needful to learn the basic concept of probability. We will cover the basics as well as the hard level problems for all levels of students for all competitive exams especially SBI PO, SBI CLERK, IBPS PO, IBPS CLERK, RRB PO, NICL AO, LIC AAO, SNAP, MAT, SSC CGL etc.

Definition:

Probability means the possibility or chances of an event occurring or happening. For example, when a coin is tossed, then we will get ahead or tail. It is a state of probability.

In an event, the happening probability is equal to the ratio of favourable outcomes to the total number of possible outcomes. It represents as,

Number of favourable outcomes = ____________________________________ Total number of possible outcomes

Sample Space:-

It is a set of all possible outcomes of an experiment. It is denoted by S. For example, the Sample space of a die, S = [ 1, 2, 3 , 4, 5, 6] The Sample space of a coin, S= [ Head, Tail]

Types of questions asked in the competitive exam:

1) Based on Coins

2) Based on Dice

3) Based on playing Cards

4) Based on Marbles or balls

5) Miscellaneous

Important Questions:

1. Question A coin is thrown two times .what is the probability that at least one tail is obtained?

A) 3/4 B) 1/4 C) 1/3 D) 2/3 E) None of these

Answer :- A Sol:

Sample space = [TT, TH, HT,HH] Total number of ways = 2 × 2 = 4. Favourite Cases = 3 P (A) = 3/4

Tricks:- P (of getting at least one tail) = 1 – P (no head)⇒ 1 – 1/4 = 3/4

2. Question What is the probability of getting a numbered card when drawn from the pack of 52 cards?

A) 1/13 B) 1/9 C) 9/13 D) 11/13 E) None of these

Answer :- C Sol: Total Cards = 52. Numbered Cards = 9 (2,3,4,5,6,7,8,9,10) in each suit Numbered cards in four suit = 4 ×9 = 36 P (E) = 36/52 = 9/13

3.Question There are 7 purple clips and 5 brown clips. Two clips are selected one by one without replacement. Find the probability that the first is brown and the second is purple.

A) 1/35 B) 35/132 C) 1/132 D) 35/144 E) None of these

Answer :- B Sol:

P (B) × P (P) = (5/12) x (7/11) = 35/132

4.Question Find the probability of getting a sum of 8 when two dice are thrown?

A) 1/8 B) 1/5 C) 1/4 D) 5/36 E) 1/3

Answer 😀 Sol: Total number of ways = 6 × 6 = 36 ways. Favorable cases = (2 , 6) (6, 2) (3, 5) (5, 3) (4, 4) — 5 ways. P (A) = 5/36 = 5/36

5.Question Find the probability of an honour card when a card is drawn at random from the pack of 52 cards.

A) 4/13 B) 1/3 C) 5/12 D) 7/52 E) None of these

Answer :-A Sol: Honor cards = 4 (A, J, Q, K) in each suit Honor cards in 4 suit = 4 × 4 = 16 P (honor card) = 16/52 = 4/13

6. Question What is the probability of a face card when a card is drawn at random from the pack of 52 cards?

A) 1/13 B) 2/13 C) 3/13 D) 4/13 E) 5/13

Answer :-C Solution: face cards = 3 (J,Q,K) in each suit Face cards in 4 suits = 3 × 4 = 12 Cards. P (face Card) = 12/52 = 3/13

7.Question If two dice are rolled together then find the probability as getting at least one ‘3’?

A) 11/36 B) 1/12 C) 1/36 D) 13/25 E) 13/36

Answer :- A Sol: Total number of ways = 6 × 6 = 36. Probability of getting number ‘3′ at least one time = 1 – (Probability of getting no number 4) = 1 – (5/6) x (5/6) = 1 – 25/36 = 11/36

8. Question If a single six-sided die is rolled then find the probability of getting either 3 or 4.

A) 1/2 B) 1/3 C) 1/4 D) 2/3 E) 1/6

Answer:- B Solution:- Total outcomes = 6 Probability of getting a single number when rolled a die = 1/6 So, P(3) = 1/6 and P(4) = 1/6 Thus, the probability of getting either 3 or 4 = P(3)+P(4) = 1/6 + 1/6 = 1/3

9. Question A container contains 1 red, 3 black, 2 pink and 4 violet gems. If a single gem is chosen at random from the container, then find the probability that it is violet or black?

A) 1/10 B) 3/10 C) 7/10 D) 9/10 E) None of these

Answer :-C Sol :- Total gems =( 1 + 3 + 2 + 4 ) = 10 probability of getting a violet gem = 4/10 Probability of getting a black gem = 3/10 Now, P ( Violet or Black) = P(violet) + P(Black) = 4/10 + 3/10 = 7/10

10.Question A jar contains 63 balls ( 1,2,3,……., 63). Two balls are picked at random from the jar one after one and without any replacement. what is the probability that the sum of both balls drawn is even?

A) 5/21 B) 3/23 C) 5/63 D) 19/63 E) None of these

Answer :- A Sol. Total balls = 63 Total even balls = 31 ( 2 , 4 , 6,……., 62) Now the required probability =³¹C₂/63C₂ = (31!/2!29!)/(63!/2!61!) = (31 × 30/1× 2)/(63×62/1×2) = (31 × 30)/(63×62) = 30/63×2 = 5/21

11.Question There are 30 students in a class, 15 are boys and 15 are girls. In the final exam, 5 boys and 4 girls made an A grade. If a student is chosen at random from the class, what is the probability of choosing a girl or an ‘A grade student?

A) 1/4 B) 3/10 C) 1/3 D) 2/3 E) None of these

Answer:- D Sol:

Here, the total number of boys = 15 and the total number of girls = 15

Also, girls getting A grade = 4 and boys getting an A grade = 5 Probability of choosing a girl = 15/30

Probability of choosing A grade student= 9/30

Now, an A-grade student chosen can be a girl. So the probability of choosing it = 4/30

Required probability of choosing a girl or an A grade student = 15/30 + 9/30 – 4/30 = 1/2 + 3/10 – 2/15 = 2/3 12. Question What is the probability when a card is drawn at random from a deck of 52 cards is either an ace or a club?

A) 2/13 B) 3/13 C) 4/13 D) 5/23 E) None of these

Answer:- C Sol: There are 4 aces in a pack, 13 club cards and 1 ace of club card.

Now, the probability of getting an ace = 4/52

Probability of getting a club = 13/52

Probability of getting an ace of club = 1/52

Required probability of getting an ace or a club

= 4/52 + 13/52 – 1/52 = 16/52 = 4/13

13. Question One card is drawn from a deck of 52 cards well shuffling. Calculate the probability that the card will not be a king.

A) 12/13 B) 3/13 C) 7/13 D) 5/23 E) None of these

Solution:

Well-shuffling ensures equally likely outcomes. Total king of a deck = 4

The number of favourable outcomes F= 52 – 4 = 48

The number of possible outcomes = 52

Therefore, the required probability

= 48/52 = 12/13

14.Question If P(A) = 7/13, P(B) = 9/13 and P(A∩B) = 4/13, find the value of P(A|B).

A) 1/9 B) 2/9 C) 3/9 D) 4/9 E) None of these

Answer :- D Solution:

P(A|B) = P(A∩B)/P(B) = (4/13)/(9/13) = 4/9.

15. Question A one rupee coin and a two rupee coin are tossed once, then calculate a sample space.

The outcomes are either Head (H) or tail(T).

Now,heads on both coins = (H,H) = HH

Tails on both coins = ( T, T) = TT

Probability of head on one rupee coin and Tail on the two rupee coins = (H, T) = HT

And Tail on one rupee coin and Head on the two rupee coin = (T, H) = TH

Thus, the sample space ,S = [HH, HT, TH, TT]

16. Question There are 20 tickets numbered 1 to 20. These tickets are mixed up and then a ticket is drawn at random. Find the probability that the ticket drawn has a number which is a multiple of 4 or 5?

A) 1/4 B) 2/13 C) 8/15 D) 9/20 E) None of these

Here, S = {1, 2, 3, 4, …., 19, 20} = 20

Multiples of 4: 4, 8, 12, 16, 20 (5 tickets) Multiples of 5: 5, 10, 15, 20 (4 tickets)

Notice that ticket number 20 is a multiple of both 4 and 5, so we have counted it twice. Therefore, we need to subtract one from the total count.

Total number of tickets with numbers that are multiples of 4 or 5: 5 + 4 – 1 = 8

The total number of tickets is 20, so the probability of drawing a ticket with a number that is a multiple of 4 or 5 is:

P = 8/20 = 2/5 = 0.4

Therefore, the probability that the ticket drawn has a number which is a multiple of 4 or 5 is 0.4 or 40%.

Direction ( 17 – 19):- In a school the total number of students is 300, 95 students like chicken only, 120 students like fish only, 80 students like mutton only and 5 students do not like anything above. If randomly one student is chosen, find the probability that

17) The student likes mutton.

18 ) he likes either chicken or mutton

19 ) he likes neither fish nor mutton.

Solution( 17-19):-

The total number of favourable outcomes = 300 (Since there are 300 students altogether).

The number of times a chicken liker is chosen = 95 (Since 95 students like chicken).

The number of times a fish liker is chosen = 120.

The number of times a mutton liker is chosen = 80.

The number of times a student is chosen who likes none of these = 5.

17. Question Find the probability that the student like mutton?

A) 3/10 B) 4/15 C) 1/10 D) 1/15 E) None of these

Answer:- B Solution:-

Therefore, the probability of getting a student who likes mutton

= 80/300 = 4/15

18. Question What is the probability that the student likes either chicken or mutton?

A) 7/12 B) 5/12 C) 3/4 D) 1/12 E) None of these

Answer:- A Solution:-

The probability of getting a student who likes either chicken or mutton = (95+80)/300 = 175/300 = 7/12

19. Question Find the probability that the student likes neither fish nor mutton.

A) 1/2 B) 1/5 C) 1/3 D) 1/4 E) 1/6

Answer:- C Solution:- The probability of getting a student who likes neither fish nor mutton = (300–120−80)/300 = 100/300 = 1/3

Direction ( 20-22):- A box contains 90 number plates numbered 1 to 90. If one number plate is drawn at random from the box then find out the probability that

20) The number is a two-digit number

21) The number is a perfect square

22) The number is a multiply of 5

20. Question Find the probability that the number is a two-digit number.

A) 1/9 B) 1/10 C) 9/10 D) 7/10 E) None of these

Answer:-C Solution : Total possible outcomes = 90 (Since the number plates are numbered from 1 to 90).

Number of favourable outcomes = 90 – 9 = 81 ( here, except 1 to 9, other numbers are two-digit number.)

Thus required probability = Number of Favourable Outcomes /Total Number of Possible Outcomes = 81/90 = 9/10.

21. Question What is the probability that the number is a perfect square?

A) 1/9 B) 1/10 C) 9/10 D) 1/7 E) None of these

Answer:- B Solution:- Total possible outcomes = 90. Number of favourable outcomes = 9 [here 1, 4, 9, 16, 25, 36, 49, 64 and 81 are the perfect squares] Thus the required probability = 9/90 =1/10

22.Question Find the probability that the number is a multiple of 5.

A) 1/5 B) 1/6 C) 1/10 D) 1/8 E) 9/10

Answer:- A Solution:- Total possible outcomes = 90. Number of favourable outcomes = 18 (here, 5 × 1, 5 × 2, 5 × 3, …., 5 × 18 are multiple of 5).

Thus, the required probability= 18/90 =1/5

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How To Solve Probability Questions Quickly

February 15, 2023

How to Solve Proabability Questions Quickly

Here , In this Page you learn Find How to Solve Probability Questions Quickly.

Formula for Probability , P(A) = \mathbf{ \frac{The Number of wanted outcomes }{The total number of Possible Outcomes}}

Therefore, probability of the occurrence of event is the number between 0 and 1 .

How to Solve Quickly Probability questions

Types 1- How to Solve Probability Questions Quickly of Random ticket or ball drawn

Question 1. Tickets numbered 1 to 20 are mixed up and then a ticket is drawn at random. What is the probability that the ticket drawn has a number which is a multiple of 3 or 5?

(a) \frac{1}{2}

(b) \frac{2}{5}

(d) \frac{9}{20}

Solution: Here, S = {1, 2, 3, 4, …., 19, 20}.

Let E = event of getting a multiple of 3 or 5 = {3, 6 , 9, 12, 15, 18, 5, 10, 20}.

P(E) = \frac{n(E)}{n(S)} = \frac{9}{20}

Correct Options (D).

Question 2 A bag contains 2 red, 3 green and 2 blue balls. Two balls are drawn at random. What is the probability that none of the balls drawn is blue?

(a) \frac{10}{21}

(b) \frac{11}{21}

(d) \frac{5}{7}

Solution: Total number of balls = (2 + 3 + 2) = 7.

Let S be the sample space.

Then, n (S) = Number of ways of drawing 2 balls out of 7.

= \frac{7 × 6}{2 × 1}

Let E = Event of drawing 2 balls, none of which is blue.

= \frac{5 × 4}{2 × 1}

P(E) = \frac{n(E)}{n(S)} = \frac{10}{21} .

Correct Options (A)

Questions 3 A bag contains 1100 tickets numbered 1, 2, 3, … 1100. If a ticket is drawn out of it at random, what is the probability that the ticket drawn has the digit 2 appearing on it?

(a) \frac{291}{1100}

(b) \frac{292}{1100}

(d) \frac{301}{1100}

Solution: Ticket has a maximum of 4 digits on it as Thousands, Hundreds Tens ,Units or TH H T U.

Number of Tickets with 2 in TH place = 0.

Number of Tickets with 2 in H place = From 200 upto 299 = 100.

Number of Tickets with 2 in T place but not in H place = 20 to 29 in T and U places and 00 to 10 except 02 in TH and H places = 10×10 = 100.

Number of Tickets with 2 ONLY in U place but not in TH H or T place

= H or T place both have (0 to 9 excluding 2) + (TH=1 & H=0 & U

= 2 & (T= 0 to 9 excluding 2) )

= (9×9) + 9 = 90.

Total Tickets with at least one 2 = 290

Probability = \frac{290}{1100} is answer.

Correct Options (c)

Type 2- How to Solve Probability Questions Quickly of boys and girls

Question 1. In a class there are 60% of girls of which 25% poor. What is the probability that a poor girl is selected is leader?

Solutions : Assume total students in the class = 100

Then Girls = 60% (100) = 60

Poor girls = 25% (60) = 15

So probability that a poor girls is selected leader = \frac{Poor girls}{ Total students} = \frac{15}{100} = 15%

Correct Options (b)

Questions 2. What is the probability that the total of two dice will be greater than 9, given that the first die is a 5?

(a) \frac{1}{3}

(b) \frac{1}{6}

(d) None of these

Solution : Let A= first die is 5

Let B = total of two dice is greater than 9

P( A ) = Possible outcomes for A and B : (5, 5), (5, 6)

P(A and B) = \frac{2}{36} = \frac{1}{18}

P(B|A) = \frac{P(A and B)}{P(A)} = \frac{1}{18} ÷ \frac{1}{6} = \frac{1}{3} .

Questions 3. If six cards are selected at random (without replacement) from a standard deck of 52 cards, what is the probability there will be no pairs? (two cards of the same denomination)

Solution: Let E i be the event that the first i cards have no pair among them. Then we want to compute P(E 6 ).

Which is actually the same as P(E 1 ∩ E 2 ∩ · · · ∩ E 6 ), since E 6 ⊂ E 5 ⊂ · · · ⊂ E 1 , implying that E 1 ∩ E 2 ∩ · · · ∩ E 6 = E 6 .

We get P(E 1 ∩ E 2 ∩ · · · ∩ E 6 ) = P(E 1 )P(E 2 |E 1 )· · · = \frac{52}{52} \frac{48}{51} \frac{44}{50} \frac{40}{49} \frac{36}{48} \frac{32}{47}

Alternatively, one can solve the problem directly using counting techniques.

Define the sample space to be (equally likely)ordered sequences of 6 cards; then, |S| = 52 · 51 · 50 · · · 47, and the event E6 has 52 · 48 · 44 · · · 32 elements

Questions 4. A group of 5 friends-Archie, Betty, Jerry, Moose, and Veronica-arrived at the movie theater to see a movie. Because they arrived late, their only seating option consists of 3 middle seats in the front row, an aisle seat in the front row, and an adjoining seat in the third row. If Archie, Jerry, or Moose must sit in the aisle seat while Betty and Veronica refuse to sit next to each other, how many possible seating arrangements are there?

Solution: Good = Total – Bad.

Total = arrangements with Archie, Jerry or Moose in the aisle seat.

Number of options for the aisle seat = 3. (Archie, Jughead, or Moose)

Number of ways to arrange the 4 other people = 4 x 3 x 2 x 1.

To combine these options, we multiply: 3 x 4 x 3 x 2 = 72.

Bad = arrangements with Archie, Jerry or Moose in the aisle seat BUT with Betty next to Veronica.

Number of options for the aisle seat = 3. (Archie, Jughead, Moose).

Number of options for the third row seat = 2. (Anyone but Betty and Veronica, since in a bad arrangement they sit next to each other.)

Number of options for the middle of the 3 remaining seats = 2. (Must be Betty or Veronica so that they sit next to each other).

Number of ways to arrange the 2 remaining people = 2 x 1.

To combine these options, we multiply: 3 x 2 x 2 x 2 = 24.

Good arrangements = 72 – 24 = 48.

Correct Option C .

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Free Mathematics Tutorials

Probability questions with solutions.

Tutorial on finding the probability of an event. In what follows, S is the sample space of the experiment in question and E is the event of interest. n(S) is the number of elements in the sample space S and n(E) is the number of elements in the event E.

Questions and their Solutions

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How to Solve Probability Questions

Probability questions are often associated with card and game problems.

How to Explain the Sum & Product Rules of Probability

Most probability questions are word problems, which require you to set up the problem and break down the information given to solve. The process to solve the problem is rarely straightforward and takes practice to perfect. Probabilities are used in mathematics and statistics and are found in everyday life, from weather forecasts to sporting events. With a little practice and a few tips, the process of calculating probabilities can be more manageable.

Find the keyword. One important tip when solving a probability word problem is to find the keyword, which helps to identify which rule of probability to use. The keywords are "and," "or" and "not." For instance, consider the following word problem: "What is the probability that Jane will choose both the chocolate and the vanilla ice cream cones given that she chooses chocolate 60 percent of the time, vanilla 70 percent of the time, and neither 10 percent of the time." This problem has the keyword "and."

Find the correct rule of probability. For problems with the keyword "and," the rule of probability to use is a multiplication rule. For problems with the keyword "or," the rule of probability to use is an addition rule. For problems with the keyword "not," the rule of probability to use is the complement rule.

Determine what event is being sought. There may be more than one event. An event is the occurrence in the problem that you are solving the probability for. The example problem is asking for the event that Jane will choose both the chocolate and the vanilla. So in essence, you want the probability of her choosing these two flavors.

Determine whether the events are mutually exclusive or independent if appropriate. When using a rule of multiplication, there are two to choose from. You use the rule P(A and B) = P(A) x P(B) when the events A and B are independent. You use the rule P(A and B) = P(A) x P(B|A) when the events are dependent. P(B|A) is a conditional probability, indicating the probability that event A occurs given that event B has already occurred. Similarly, for the rules of addition, there are two to choose from. You use the rule P(A or B) = P(A) + P(B) if the events are mutually exclusive. You use the rule P(A or B) = P(A) + P(B) - P(A and B) when the events are not mutually exclusive. For the complement rule, you always use the rule P(A) = 1 - P(~A). P(~A) is the probability that event A does not occur.

Find the separate parts of the equation. Each equation of probability has different parts that need to be filled to solve the problem. For the example, you determined the keyword is "and," and the rule to use is a rule of multiplication. Because the events are not dependent, you will use the rule P(A and B) = P(A) x P(B). This step sets P(A) = probability of event A occurring and P(B) = probability of event B occurring. The problem says that P(A = chocolate) = 60% and P(B = vanilla) = 70%.

Substitute the values into the equation. You can substitute the word "chocolate" when you see the event A and the word "vanilla" when you see the event B. Using the appropriate equation for the example and substituting the values, the equation is now P(chocolate and vanilla) = 60% x 70%.

Solve the equation. Using the previous example, P(chocolate and vanilla) = 60 percent x 70 percent. Breaking down the percentages into decimals will yield 0.60 x 0.70, found by dividing both percentages by 100. This multiplication results in the value 0.42. Converting the answer back to a percentage by multiplying by 100 will yield 42 percent.

Two events are known to be mutually exclusive if they both cannot occur at the same time. If they can occur at the same time, they are not. Two events are known to be independent if one event does not depend on the outcome of the other event. These definitions are used to help complete the previous steps; a working knowledge of these is required to solve these problems.

How to calculate conditional probabilities, how to calculate probability with percentages, methods of probability, how to calculate discrete probability distribution, tricks to solving percentage word problems, how to calculate a recurrence interval, how to solve inequalities with fractions, how to read a weather forecast, how to set up a pareto chart in minitab, types of probability math questions, sat math prep ii: exponents, ratios and percentages, what are subtraction sentences, how to calculate probability and normal distribution, how to calculate mode in minitab, the law of probability, how to calculate binomial probability.

About the Author

Michelle Friesen began writing in 2003. Contributing to eHow, she is also a software engineer and adjunct instructor of statistics and computer information systems. Friesen holds a Master of Science in engineering management and a certificate in financial engineering, as well as Bachelor of Science degrees in applied mathematics and computer science from the Missouri University of Science and Technology.

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Solve probability, percentile, and more

uppose that a category of world class runners are known to run a marathon (26 miles) in an average of 147 minutes with a standard deviation of 15 minutes. Consider 49 of the races.

= the average of the 49 races.

a. Give the distribution of X .

(Round your standard deviation to two decimal places.)

b.Find the probability that the average of the sample will be between 146 and 149 minutes in these 49 marathons. (Round your answer to four decimal places.)

c. Find the 60th percentile for the average of these 49 marathons. (Round your answer to two decimal places.)

d. Find the median of the average running times.

1 Expert Answer

Daniel K. answered • 5d

MS in Statistics with 5+ years of tutoring experience

According to the central limit theorem the sample mean of 49 races should be approximately normally distributed with a mean of 147 with a sd of 15/sqrt(7) = 15/7 . Now you can use Z scores and the Z table to solve the rest:

P(146<X<149) = P( (146-147)/(15/7) < Z < (149-147)/(15/7) ) go from here/. LMK if you need more help

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Statistics and probability

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15 Probability Questions And Practice Problems (KS3 & KS4): Harder GCSE Exam Style Questions Included

Beki Christian

Probability questions and probability problems require students to work out how likely it is that something is to happen. Probabilities can be described using words or numbers. Probabilities range from 0 to 1 and can be written as fractions, decimals or percentages.

Here you’ll find a selection of probability questions of varying difficulty showing the variety you are likely to encounter in KS3 and KS4 including several GCSE exam style questions.

What are some real life examples of probability?

How to calculate probabilities, year 7 probability questions, year 8 probability questions, year 9 probability questions, year 10 probability questions, gcse foundation probability questions, gcse higher probability questions, looking for more probability questions and resources, looking for more ks3 and ks4 maths questions.

Free GCSE maths revision resources for schools In addition to the GCSE maths tuition we provide to secondary schools more support is available free for GCSE maths revision including: – GCSE maths past papers – GCSE maths worksheets – GCSE maths questions

The more likely something is to happen, the higher its probability. We think about probabilities all the time. For example, you may have seen that there is a 20% chance of rain on a certain day or thought about how likely you are to roll a 6 when playing a game, or to win in a raffle when you buy a ticket.

The probability of something happening is given by:

We can also use the following formulae to help us calculate probabilities and solve problems:

KS3 probability questions

In KS3 probability questions introduce the idea of the probability scale and the fact that probabilities sum to one. We look at theoretical and experimental probability as well as learning about sample space diagrams and venn diagrams.

1. Which number could be added to this spinner to make it more likely that the spinner will land on an odd number than a prime number?

Currently there are two odd numbers and two prime numbers so the chances of landing on an odd number or a prime number are the same. By adding 3, 5 or 11 you would be adding one prime number and one odd number so the chances would remain equal.

By adding 9 you would be adding an odd number but not a prime number. There would be three odd numbers and two prime numbers so the spinner would be more likely to land on an odd number than a prime number.

2. Ifan rolls a fair dice, with sides labelled A, B, C, D, E and F. What is the probability that the dice lands on a vowel?

A and E are vowels so there are 2 outcomes that are vowels out of 6 outcomes altogether.

Therefore the probability is \frac{2}{6} which can be simplified to \frac{1}{3} .

3. Max tested a coin to see whether it was fair. The table shows the results of his coin toss experiment:

Heads Tails

26 41

What is the relative frequency of the coin landing on heads?

Max tossed the coin 67 times and it landed on heads 26 times.

\text{Relative frequency (experimental probability) } = \frac{\text{number of successful trials}}{\text{total number of trials}} = \frac{26}{67}

4. Grace rolled two dice. She then did something with the two numbers shown. Here is a sample space diagram showing all the possible outcomes:

What did Grace do with the two numbers shown on the dice?

Add them together

Subtract the number on dice 2 from the number on dice 1

Multiply them

Subtract the smaller number from the bigger number

For each pair of numbers, Grace subtracted the smaller number from the bigger number.

For example, if she rolled a 2 and a 5, she did 5 − 2 = 3.

5. Alice has some red balls and some blue balls in a bag. Altogether she has 25 balls. Alice picks one ball from the bag. The probability that Alice picks a red ball is x and the probability that Alice picks a blue ball is 4x. Work out how many blue balls are in the bag.

Since the probability of mutually exclusive events add to 1:

\begin{aligned} x+4x&=1\\\\ 5x&=1\\\\ x&=\frac{1}{5} \end{aligned}

\frac{1}{5} of the balls are red and \frac{4}{5} of the balls are blue.

6. Arthur asked the students in his class whether they like maths and whether they like science. He recorded his results in the venn diagram below.

How many students don’t like science?

We need to look at the numbers that are not in the ‘Like science’ circle. In this case it is 9 + 7 = 16.

KS4 probability questions

In KS4 probability questions involve more problem solving to make predictions about the probability of an event. We also learn about probability tree diagrams, which can be used to represent multiple events, and conditional probability.

One of the first probability slides on tree diagrams for GCSE students on Third Space Learning's online intervention.

7. A restaurant offers the following options:

Starter – soup or salad

Main – chicken, fish or vegetarian

Dessert – ice cream or cake

How many possible different combinations of starter, main and dessert are there?

The number of different combinations is 2 × 3 × 2 = 12.

8. There are 18 girls and 12 boys in a class. \frac{2}{9} of the girls and \frac{1}{4} of the boys walk to school. One of the students who walks to school is chosen at random. Find the probability that the student is a boy.

First we need to work out how many students walk to school:

\frac{2}{9} \text{ of } 18 = 4

\frac{1}{4} \text{ of } 12 = 3

7 students walk to school. 4 are girls and 3 are boys. So the probability the student is a boy is \frac{3}{7} .

9. Rachel flips a biased coin. The probability that she gets two heads is 0.16. What is the probability that she gets two tails?

We have been given the probability of getting two heads. We need to calculate the probability of getting a head on each flip.

Let’s call the probability of getting a head p.

The probability p, of getting a head AND getting another head is 0.16.

Therefore to find p:

The probability of getting a head is 0.4 so the probability of getting a tail is 0.6.

The probability of getting two tails is 0.6 × 0.6 = 0.36 .

10. I have a big tub of jelly beans. The probability of picking each different colour of jelly bean is shown below:

If I were to pick 60 jelly beans from the tub, how many orange jelly beans would I expect to pick?

First we need to calculate the probability of picking an orange. Probabilities sum to 1 so 1 − (0.2 + 0.15 + 0.1 + 0.3) = 0.25.

The probability of picking an orange is 0.25.

The number of times I would expect to pick an orange jelly bean is 0.25 × 60 = 15 .

11. Dexter runs a game at a fair. To play the game, you must roll a dice and pick a card from a deck of cards.

To win the game you must roll an odd number and pick a picture card. The game can be represented by the tree diagram below.

Dexter charges players £1 to play and gives £3 to any winners. If 260 people play the game, how much profit would Dexter expect to make?

Completing the tree diagram:

Probability of winning is \frac{1}{2} \times \frac{4}{13} = \frac{4}{26}

If 260 play the game, Dexter would receive £260.

The expected number of winners would be \frac{4}{26} \times 260 = 40

Dexter would need to give away 40 × £3 = £120 .

Therefore Dexter’s profit would be £260 − £120 = £140.

12. A coin is tossed three times. Work out the probability of getting two heads and one tail.

There are three ways of getting two heads and one tail: HHT, HTH or THH.

The probability of each is \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} = \frac{1}{8}

Therefore the total probability is \frac{1}{8} +\frac{1}{8} + \frac{1}{8} = \frac{3}{8}

13. 200 people were asked about which athletics event they thought was the most exciting to watch. The results are shown in the table below.

A person is chosen at random. Given that that person chose 100m, what is the probability that the person was female?

Since we know that the person chose 100m, we need to include the people in that column only.

In total 88 people chose 100m so the probability the person was female is \frac{32}{88} .

14. Sam asked 50 people whether they like vegetable pizza or pepperoni pizza.

37 people like vegetable pizza.

25 people like both.

3 people like neither.

Sam picked one of the 50 people at random. Given that the person he chose likes pepperoni pizza, find the probability that they don’t like vegetable pizza.

We need to draw a venn diagram to work this out.

We start by putting the 25 who like both in the middle section. The 37 people who like vegetable pizza includes the 25 who like both, so 12 more people must like vegetable pizza. 3 don’t like either. We have 50 – 12 – 25 – 3 = 10 people left so this is the number that must like only pepperoni.

There are 35 people altogether who like pepperoni pizza. Of these, 10 do not like vegetable pizza. The probability is \frac{10}{35} .

15. There are 12 marbles in a bag. There are n red marbles and the rest are blue marbles. Nico takes 2 marbles from the bag. Write an expression involving n for the probability that Nico takes one red marble and one blue marble.

We need to think about this using a tree diagram. If there are 12 marbles altogether and n are red then 12-n are blue.

To get one red and one blue, Nico could choose red then blue or blue then red so the probability is:

Third Space Learning’s free GCSE maths resource library contains detailed lessons with step-by-step instructions on how to solve ratio problems, as well as worksheets with ratio and proportion practice questions and more GCSE exam questions.

Take a look at the probability lessons today – more are added every week.

Do you have students who need extra support in maths? Every week Third Space Learning’s maths specialist tutors support thousands of students across hundreds of schools with weekly online 1-to-1 lessons and maths interventions designed to plug gaps and boost progress. Since 2013 we’ve helped over 150,000 primary and secondary students become more confident, able mathematicians. Find out more about our GCSE Maths tuition or request a personalised quote for your school to speak to us about your school’s needs and how we can help.

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Download this 15 Probability Questions And Practice Problems (KS3 & KS4) Worksheet

Help your students prepare for their Maths GSCE with this free Probability worksheet of 15 multiple choice questions and answers.

FREE GCSE Circle Theorems Worksheet

Help your students prepare for their maths GCSE with this free circle theorems worksheet of 44 questions and answers.

Contains 36 skills-based questions, 4 applied and 4 foundation and higher level GCSE exam style questions.

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How to solve Probability questions? [closed]

I am preparing GRE Test which has Math Part.

How to solve probability questions in a real quick ?

One marble is randomly selected from a bag that contains only 4 black marbles, 3 red marbles, 5 yellow marbles, and 4 green marbles.

Quantity a) The probability of selecting either a black marble or a red marble Quantity b) The probability of selecting either a yellow marble or a green marble

Whether Quantity a will be greater ? Whether Quantity b will be greater ? Whether two Quantities are equal ? Relationship cannot be determined?

probability

$\begingroup$ Are you sure you describe the question correctly? Do you mean "4 black marbles"? $\endgroup$ – aelguindy Mar 1, 2012 at 21:32

3 Answers 3

I assume you mean the bag contains 4 black marbles.

There are 16 marbles total and we assume that outcomes are equally likely. That is, we assume that any particular marble is just as likely to be chosen as any other. Then, since there are 16 marbles total, the probability of selecting any particular marble is $1/16$.

In general, if outcomes are equally likely, then to find the probability of an event $A$, compute $$\text { size of }A\over\text{total number of outcomes}.$$ In the above "size of $A$" is the number of outcomes that make up $A$.

For $A$ being the event "a black or red marble was chosen", the size is $7$, (one of the 4 black or one of the 3 red were chosen). So, the probability that a black or red marble was chosen is $ 7/16$.

If $A$ is the event the event "a yellow or green marble is chosen", the size of $A$ is 9, and the corresponding probability is $9/16$.

Obviously, the second quantity is the greater one (which could have been seen without doing any arithmetic, since the size of "yellow or green" is larger than the size of "black or red").

$\begingroup$ Thank you !! Yes it is 4 Black marbles and answer is correct, quantity b is greater !! And thank you for solving the question. This helps me to solve similar problems $\endgroup$ – goofyui Mar 3, 2012 at 23:53

I am assuming you meant the bag "contains 4 black marbles, ..."

You select one marble out of 16 marbles. 7 marbles are red or black, therefore the probability that your pick is red or black is $7/16$.

The probability that it is yellow or green can be calculated either by counting the number of yellow and green marbles, in this case 9 and dividing by 16.

Or you can notice that the probability of being red or black and that of being yellow and green sum to one. Therefore, the probabolity you get yellow or green is $1 - 7/16 = 9/16$.

$\begingroup$ yes i missed to mention 4 black marbles. But your definition doesnt say, whether Quantity A is greater or B is Greater or two quantities are equal or Relationship Undetermined !!! $\endgroup$ – goofyui Mar 4, 2012 at 0:00

There's no algorithm (ie solution machine) that you can use to solve probability questions. You have to think about each question.

Most GRE questions are about equally likely outcomes. The probability is the number of outcomes favourable to the question, divided by the total number of possible outcomes.

In your question: there are 18 marbles, so there are 18 possible outcomes. For (a) Favourable outcomes are black or red. There are 7 such outcomes, so probability is 7/18 For (b) Favourable outcomes are yellow or green. There are 9 such outcomes, so probability is 9/18

Not the answer you're looking for? Browse other questions tagged probability .

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Probability Questions

The probability questions , with answers, are provided here for students to make them understand the concept in an easy way. The chapter Probability has been included in Class 9, 10, 11 and 12. Therefore, it is a very important chapter. The questions here will be provided, as per NCERT guidelines. Get Probability For Class 10 at BYJU’S.

The application of probability can be seen in Maths as well as in day to day life. It is necessary to learn the basics of this concept. The questions here will cover the basics as well as the hard level problems for all levels of students. Thus, students will be confident in solving problems based on it. Also, solving these probability problems will help them to participate in competitive exams, going further.

Definition: Probability is nothing but the possibility of an event occurring. For example, when a test is conducted, then the student can either get a pass or fail. It is a state of probability.

Probability Questions & Answers

1. Two coins are tossed 500 times, and we get:

Two heads: 105 times

One head: 275 times

No head: 120 times

Find the probability of each event to occur.

Solution: Let us say the events of getting two heads, one head and no head by E 1 , E 2 and E 3 , respectively.

P(E 1 ) = 105/500 = 0.21

P(E 2 ) = 275/500 = 0.55

P(E 3 ) = 120/500 = 0.24

The Sum of probabilities of all elementary events of a random experiment is 1.

P(E 1 )+P(E 2 )+P(E 3 ) = 0.21+0.55+0.24 = 1

2. A tyre manufacturing company kept a record of the distance covered before a tyre needed to be replaced. The table shows the results of 1000 cases.

If a tyre is bought from this company, what is the probability that :

(i) it has to be substituted before 4000 km is covered?

(ii) it will last more than 9000 km?

(iii) it has to be replaced after 4000 km and 14000 km is covered by it?

Solution: (i) Total number of trials = 1000.

The frequency of a tyre required to be replaced before covering 4000 km = 20

So, P(E 1 ) = 20/1000 = 0.02

(ii) The frequency that tyre will last more than 9000 km = 325 + 445 = 770

So, P(E 2 ) = 770/1000 = 0.77

(iii) The frequency that tyre requires replacement between 4000 km and 14000 km = 210 + 325 = 535.

So, P(E 3 ) = 535/1000 = 0.535

3. The percentage of marks obtained by a student in the monthly tests are given below:

Based on the above table, find the probability of students getting more than 70% marks in a test.

Solution: The total number of tests conducted is 5.

The number of tests when students obtained more than 70% marks = 3.

So, P(scoring more than 70% marks) = ⅗ = 0.6

4. One card is drawn from a deck of 52 cards, well-shuffled. Calculate the probability that the card will

(i) be an ace,

(ii) not be an ace.

Solution: Well-shuffling ensures equally likely outcomes.

(i) There are 4 aces in a deck.

Let E be the event the card drawn is ace.

The number of favourable outcomes to the event E = 4

The number of possible outcomes = 52

Therefore, P(E) = 4/52 = 1/13

(ii) Let F is the event of ‘card is not an ace’

The number of favourable outcomes to F = 52 – 4 = 48

Therefore, P(F) = 48/52 = 12/13

5. Two players, Sangeet and Rashmi, play a tennis match. The probability of Sangeet winning the match is 0.62. What is the probability that Rashmi will win the match?

Solution: Let S and R denote the events that Sangeeta wins the match and Reshma wins the match, respectively.

The probability of Sangeet to win = P(S) = 0.62

The probability of Rashmi to win = P(R) = 1 – P(S)

= 1 – 0.62 = 0.38

6. Two coins (a one rupee coin and a two rupee coin) are tossed once. Find a sample space.

Solution: Either Head(H) or Tail(T) can be the outcomes.

Heads on both coins = (H,H) = HH

Head on 1st coin and Tail on the 2nd coin = (H,T) = HT

Tail on 1st coin and Head on the 2nd coin = (T,H) = TH

Tail on both coins = (T,T) = TT

Therefore, the sample space is S = {HH, HT, TH, TT}

7. Consider the experiment in which a coin is tossed repeatedly until a head comes up. Describe the sample space.

Solution: In the random experiment where the head can appear on the 1st toss, or the 2nd toss, or the 3rd toss and so on till we get the head of the coin. Hence, the required sample space is :

S= {H, TH, TTH, TTTH, TTTTH,…}

8. Consider the experiment of rolling a die. Let A be the event ‘getting a prime number’, B be the event ‘getting an odd number’. Write the sets representing the events

(ii) A and B

(iii) A but not B

(iv) ‘not A’.

Solution: S = {1, 2, 3, 4, 5, 6}, A = {2, 3, 5} and B = {1, 3, 5}

(i) A or B = A ∪ B = {1, 2, 3, 5}

(ii) A and B = A ∩ B = {3,5}

(iii) A but not B = A – B = {2}

(iv) not A = A′ = {1,4,6}

9. A coin is tossed three times, consider the following events.

P: ‘No head appears’,

Q: ‘Exactly one head appears’ and

R: ‘At Least two heads appear’.

Check whether they form a set of mutually exclusive and exhaustive events.

Solution: The sample space of the experiment is:

S = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT} and

Q = {HTT, THT, TTH},

R = {HHT, HTH, THH, HHH}

P ∪ Q ∪ R = {TTT, HTT, THT, TTH, HHT, HTH, THH, HHH} = S

Therefore, P, Q and R are exhaustive events.

P ∩ R = φ and

Therefore, the events are mutually exclusive.

Hence, P, Q and R form a set of mutually exclusive and exhaustive events.

10. If P(A) = 7/13, P(B) = 9/13 and P(A∩B) = 4/13, evaluate P(A|B).

Solution: P(A|B) = P(A∩B)/P(B) = (4/13)/(9/13) = 4/9.

Video Lesson

Probability important topics.

Probability Important Questions

Practice Questions

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Please help me solve 22; How do you set up the PMF for this...

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Please help me solve 22; How do you set up the PMF for this problem, it's confusing me

Answer & Explanation

21) Distribution of X : Binomial(5, 0.1)

22) The mean of X is 0.50.

21) Let X be a random variable that represents the number of sugar-addicted among 5 people.

Let's consider "finding a sugar-addicted person" as success.

Probability of success (p) = 0.10

Number of trials (n) = 5

Since number of trials are finite, each trial can result in only two mutually exclusive outcomes (success and failure), probability of success remains constant in each of the trials and outcomes in each trial are independent of others, therefore we can consider that X is following binomial distribution with parameters n = 5 and p = 0.10.

X ~ Binomial(5, 0.1)

The PMF of X is given by,

P ( X = x ) = ( x n ) ( p ) x ( 1 − p ) n − x , x = 0 , 1 , . . . , 5

22) The mean of a binomial distribution is given by,

M e a n = n p

M e a n = 5 × 0 . 1 0

M e a n = 0 . 5 0

The mean of X is 0.50.

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