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Probability with replacement – Explanation & Examples

Probability with replacement and independence:

Finding probabilities of some related events:, formula for probability with replacement:, tree diagram for solving probability with replacement:, a few more examples:, practice questions, answer key:, probability with replacement – explanation & examples.

Probability with replacement title

After reading this article, you should be able to:

What does probability with replacement mean

To understand what “ probability with replacement ” means, let’s start with an example. Suppose we have a box that contains 3 orange balls and 2 blue balls. We are asked to draw two balls, one at a time. Now, what is the probability that both balls are orange?

Before we answer this question, note that there are two ways we can perform this experiment:

Method 1 (With replacement):

Probability with replacement first ball draw

Method 2 (Without replacement):

Probability with replacement 2nd ball draw

From the above example, we note that “ probability with replacement ” refers to calculating such events’ probabilities. We have a collection of objects, and we sample from the collection k-times. Each time we sample, we replace the sampled object in the collection.

In probability theory, two events are said to be independent if one event’s outcome does not affect the probability of the other event. For example, if I toss a coin two times, the first toss (Head or Tail) outcome does not affect the probability of the outcome of the second toss. Similarly, the outcome of the second toss does not affect the first toss in any manner. Hence, these two events are said to be independent.

The draws in probability with replacement are independent events . For instance, we note from the example above that we were doing the first draw, and there were 3 orange and 2 blue balls in the box. When we were doing the second draw, again, there were 3 orange and 2 blue balls in the box. The first draw did not affect the second draw and vice versa.

In contrast, in Method 2, i.e., without replacement, the first draw will change the number of either the orange or the blue balls. So the outcome of the second draw is dependent on the first draw when sampling without replacement .

How to calculate Probability with replacement:

Now let’s try to answer the question we asked in the example above, i.e., what is the probability that both balls are orange? To answer that, we first need to know the probability of drawing a orange ball in each draw.

Event1 = First ball is orange, and

Event2 = Second ball is orange

As there are 3 orange balls (let’s call them O1, O2, O3) and 2 blue balls (let’s call them B1 and B2) and we are equally likely to draw any one of them, hence

$P(\textrm{Event1}) = \textrm{number of orange balls}/ \textrm{total number of balls}$

$\qquad \qquad \quad= 3/5$.

In the second draw, we again have three 3 orange and 2 blue balls, so

$P(\textrm{Event2}) = 3/5$.

Remember that when two events are independent, then $P(\textrm{Event1 and Event2}) = P(\textrm{Event1}) \times P(\textrm{Event2})$. So the probability that both balls are orange is given as

$P(\textrm{Event1 and Event2}) = 3/5 \times 3/5 = 9/25$.

No orange balls

Let us suppose we are interested in finding the probability that no orange ball is drawn. We know from basic probability theory that if an event’s probability is P, then the probability that the event does not occur is 1-P. So,

$ P(\textrm{first ball is not orange}) = 1 – P( \textrm{first ball is orange}) = 1 – 3/5 = 2/5$.

$P(\textrm{Second ball is not orange}) = 1 – P( \textrm{Second ball is orange}) = 1 – 3/5 = 2/5$.

Finally, due to replacement, both draws are independent and hence

$P(\textrm{Both balls are not orange}) = 2/5 \times 2/5 = 4/25$.

At-least one orange ball:

Another related example is to find the probability of drawing at-least one orange ball. Using the above notation, we are interested in P(Event1 or Event2). Note that or in this context is the logical-OR which means either Event1 or Event2 or both. From basic axioms of probability, we know that

$P(\textrm{Event1 or Event2}) = P(\textrm{Event1}) + P(\textrm{Event2}) – P(\textrm{Event1 and Event2})$

$= 3/5 + 3/5 – 9/25$

Probability with replacement appears in various forms, and there is no simple formula that applies to all situations. To keep the discussion simple, we describe formulas for a simple example scenario.

Let us suppose, we have a collection of 2 different items. Let’s call them $c_1$ and $c_2$. Let $c_1$ appears $i$-times and $c_2$ appears $j$-times, in the collection. We sample the collection $k$-times. Let $N=i+j$.

$P(c_1 \textrm{ appears k-times}) = \left(\frac{i}{N}\right)^k$.

$P(c_1 \textrm{ does not appear in any draw}) = \left(1 – \frac{i}{N}\right)^k = \left(\frac{j}{N}\right)^k = P(c_2 \textrm{ appears k-times})$.

$P(c_1 \textrm{ appeasrs atleast once}) = 1-\left (1 – \frac{i}{N}\right)^k$.

We use Tree diagrams to organize information. These can be used to solve probability problems by representing all possible events and their respective outcomes. The probability of an event is written on its branch, and to calculate the overall probability, we multiply the probabilities along the branches. Let’s consider an example.

Let’s suppose there are thirteen balls in a box. Five balls are Green(G), and eight balls are Red(R). If we draw two balls, one at a time, with replacement, find the probability of the following events:

We can solve this question by drawing a tree diagram as shown below:

Probability with replacement third diagram

Let’s consider some more examples to clarify the concept of probability with replacement further.

Four cards are picked randomly, with replacement, from a regular deck of 52 playing cards. Find the probability that all four are aces.

There are four aces in a deck, and as we are replacing after each sample, so

$P(\textrm{First Ace}) = P(\textrm{Second Ace}) = P(\textrm{Third Ace}) = P(\textrm{Fouth Ace}) = \frac{4}{52}$.

All four samples are independent, so

$P(\textrm{all four are aces}) = \frac{4}{52} \times \frac{4}{52} \times \frac{4}{52} \times \frac{4}{52}=\frac{1}{28561}$.

A bag contains 3 books named Maths(M), Science(S), and Physics(P). A book is chosen at random, and after recording its name, it is returned to the bag. If this event occurred five times, find the probability of the following:

1. There are six possible permutations in which three books can be sampled without any book being selected twice, i.e., $\{\textrm{M}, \textrm{S}, \textrm{P}\}$, $\{\textrm{M}, \textrm{P}, \textrm{S}\}$, $\{\textrm{S}, \textrm{M}, \textrm{P}\}$, $\{\textrm{S}, \textrm{P}, \textrm{M}\}$, $\{\textrm{P}, \textrm{S}, \textrm{M}\}$, $\{\textrm{P}, \textrm{M}, \textrm{S}\}$. The probability of each permutation is the same so we show the calculation of the probability of $\{\textrm{M}, \textrm{S}, \textrm{P}\}$ only.

$P(\textrm{First book is Maths}) = \frac{1}{3}$

$P(\textrm{Second book is Science}) = \frac{1}{3}$

$P(\textrm{Third book is Physics}) = \frac{1}{3}$

Due to replacement, the probability of drawing each book is the same, and all draws are independent and so

$P(\{\textrm{M}, \textrm{S}, \textrm{P}\}) = \frac{1}{3} \times \frac{1}{3} \times \frac{1}{3} = \frac{1}{27}$.

Since there are six possibilities and each event is disjoint from the other, so

$P(\textrm{Each book is selected once}) = 6 \times \frac{1}{27} = \frac{2}{9}$.

$P(\textrm{First book is Science}) = \frac{1}{3}$

$P(\textrm{Second book is Science}) = P(\textrm{Third book is Science}) = \frac{1}{3}$

$P(\textrm{Science is selected three times}) = \frac{1}{3} \times \frac{1}{3} \times \frac{1}{3} = \frac{1}{29}$.

3. Let us define:

Event1: The first book is not Maths

Event2: Second book is not Maths

Event3: Third book is not Maths

$P(\textrm{Math is not selected}) = P(\textrm{Event1 and Event2 and Event3})$. From basic probability theory,

$P(\textrm{First book is not Maths}) = 1 – P(\textrm{First book is Maths})$, so

$P(\textrm{First book is not Maths}) = 1 – \frac{1}{3} = \frac{2}{3}$.

$P(\textrm{Second book is not Maths}) = P(\textrm{Third book is not Maths}) = \frac{2}{3}$.

As discussed earlier, the draws in sampling by replacement are independent and so

$P(\textrm{Event1 and Event2 and Event3}) = P(\textrm{Event1}) \times P(\textrm{Event2}) \times P(\textrm{Event3})$.

$P(\textrm{Maths is not selected}) = \frac{2}{3} \times \frac{2}{3} \times \frac{2}{3} = \frac{8}{27}$.

1 .Two cards are picked randomly, with replacement, from a regular deck of 52 playing cards. 12 cards are court cards(C), and 40 cards are spot cards(S). Draw a tree diagram to list all possible outcomes and their corresponding probabilities.

2 .The letters of the word ‘SUCCESS’ are printed on 7 cards. Jacob chooses a card at random, replaces it, then chooses a card again. Calculate the probability that only one of the cards he chooses has the letter C printed on it.

3 .Two cards are picked randomly, with replacement, from a regular deck of 52 playing cards. 4 cards are Kings, and 4 cards are Queens in a deck. What is the probability that one King and one Queen are chosen?

Probability with replacement treediagram

2. C’ represents Not the letter C.

probabilitywithreplacement example

For one of the card he chooses has ‘ C ‘ printed on it: $(2/7 \times 5/7)+(5/7 \times2/7)=20/49$.

3. $P(\textrm{one King and one Queen}) = P(\textrm{1st King and 2nd Queen}) + P(\textrm{1st Queen and 2nd King})$

$\qquad \qquad \qquad \qquad \qquad \qquad \quad = \frac{4}{52}\times\frac{4}{52} + \frac{4}{52}\times\frac{4}{52}$

$\qquad \qquad \qquad \qquad \qquad \qquad \quad = \frac{2}{169}$

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Probability with Replacement

Probability with Replacement is used for questions where the outcomes are returned to the sample space again. This means that once the item is selected, it is replaced in the sample space, so the number of elements of the sample space remains unchanged.

A jar contains five balls numbered \( 1, 2, 3, 4 \) and \( 5 \). A ball is chosen at random, and its number is recorded. The ball is then returned to the jar. This is done a total of five times.

(a) Find the probability that each ball is selected exactly once.

\( \begin{aligned} \displaystyle &=\frac{5}{5} \times\frac{4}{5} \times\frac{3}{5} \times\frac{2}{5} \times\frac{1}{5} \\ &= \frac{4!}{5^4} \end{aligned} \)

(b) Find the probability that at least one ball is not selected.

\( \begin{aligned} \displaystyle &= 1-\frac{4!}{5^4} \end{aligned} \)

(c) Find the probability that exactly one of the balls is not selected.

\( \begin{aligned} \displaystyle \Pr(\text{Ball 1 is not selected and all the rest at least once}) &= \frac{4}{5} \times \frac{4}{5} \times \frac{3}{5} \times \frac{2}{5} \times \frac{1}{5} \\ &= 4 \times \frac{4!}{5^5} \\ \Pr(\text{Exactly one not selected}) &= 5 \times 4 \times \frac{4!}{5^5} \\ &= 4 \times \frac{4!}{5^4} \end{aligned} \)

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Probability problems and sampling with and without replacement

Theoretical probability, simulation and the tools of probability, a note on playing cards.

pROBABILITY WITH REPLACEMENT

Probability basics, probability without replacement, house of cards activity using probability without replacement.

how to solve probability problems with replacement

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Probability with replacement marbles

Two marbles are drawn at random and with replacement from a box containing $2$ red, $3$ green, and $4$ blue marbles. Let's define the following events: A={two red marbles are drawn} B={ two green marbles are drawn} C={two blue marbles are drawn}.

Let's say i want to find the probability of A. Since it's with replacement the first time i'm drawing, the probability would be $\frac29$ and the second time would also be $\frac29$ which would be $\frac4{81}$ . Is this the correct way of thinking this?

probability

3 Answers 3

Yes, you are on a right track:

Total number of balls always remains $9$ .

For event $A$ : There are $2$ Red balls, for both draws: $$P(A)=\frac29\cdot \frac29=\frac4{81}$$ For event $B$ : There are $3$ Green Balls, for both draws: $$P(B)=\frac39\cdot\frac39=\frac{9}{81}$$ For event $C$ : There are 4 Blue Balls, for both draws: $$P(C)=\frac49\cdot \frac49=\frac{16}{81}$$

$P(A)=(\frac29)\cdot(\frac29)$ $P(B)=(\frac39)\cdot(\frac39)$ $P(C)=(\frac49)\cdot(\frac49)$

$\begingroup$ use Mathjax formatting: math.meta.stackexchange.com/questions/5020/… $\endgroup$ – pooja somani Nov 12, 2018 at 6:14

There is a very simple equation that everyone seems to forget to mention: $(x/t)^n$ where $x$ is the number of desired objects, $t$ is the total amount of objects, and $n$ is the number of times that you are drawing the object. For $P(A)$ it would be as follows: $(2/9)^2$ ; $P(B)$ would be: $(3/9)^2$ or $(1/3)^2$ ; etc.

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GRE Math : Outcomes

Study concepts, example questions & explanations for gre math, all gre math resources, example questions, example question #1 : how to find the probability of an outcome.

A jar contains 10 red marbles, 4 white marbles, and 2 blue marbles. Two are drawn in sequence, not replacing after each draw.

The probability of drawing two red marbles

The probability of drawing exactly one blue marble.

The quantities are equal.

Quantity B is greater.

Quantity A is greater.

The relationship cannot be determined from the information given.

Note that there are 16 total marbles. A is simply a set of sequential events. On the first, you have 10/16 chances to draw a red. Supposing this red is not replaced, the chance of drawing a second red will be 9/15; therefore, the probability of A is (10/16) * (9/15) = 0.375. Event B is translated into 2 events: Blue + (White or Red) or (White or Red) + Blue. The probabilities of each of these events, added together would be (2/16) * (14/15) + (14/16) * (2/15) = 0.2333333333; therefore, A is more probable.

Example Question #2 : How To Find The Probability Of An Outcome

In a bowl containing 10 marbles, 5 are blue and 5 are pink. If 2 marbles are picked randomly, what is the probability that the 2 marbles will not both be pink?

To solve this question, you can solve for the probability of choosing 2 marbles that are pink and subtracting that from 1 to obtain the probability of selecting any variation of marbles that are not both pink.

The probability of picking 2 marbles that are both pink would be the product of the probability of choosing the first pink marble multiplied by the probability of choosing a second pink marble from the remaining marbles in the mix.

This would be 1/2 * 4/9 = 2/9.

To obtain the probability that is asked, simply compute 1 – (2/9) = 7/9.

The probability that the 2 randomly chosen marbles are not both pink is 7/9.

Example Question #1 : Outcomes

Choose a number at random from 1 to 5.

The probability of choosing an even number

The probability of choosing an odd number

Cannot be determined

Column A is greater

Column B is greater

Column A and B are equal

There are two even numbers and three odd numbers, so P (even) = 2/5 and P (odd) = 3/5.

Example Question #4 : How To Find The Probability Of An Outcome

Two fair dice are thrown. What is the probability that the outcome will either total 7 or include a 3?

If a die is rolled twice, there are 6 * 6 = 36 possible outcomes.

Each number is equally probable in a fair die. Thus you only need to count the number of outcomes that fulfill the requirement of adding to 7 or including a 3. These include:

This is 15 possibilities. Thus the probability is 15/36 = 5/12.

Box A has 10 green balls and 8 black balls.

Box B has 9 green balls and 5 black balls.

What is the probability if one ball is drawn from each box that both balls are green?

Note that drawing balls from each box are independent events. Thus their probabilities can be combined with multiplication.

Probability of drawing green from A:

Probability of drawing green from B:

5 / 9 * 9 / 14 = 5 / 14

The probability that events A and/or B will occur is 0.88.

Quantity A: The probability that event A will occur.

Quantity B: 0.44.

The two quantities are equal.

The only probabilites that we know from this is that P(only A) + P(only B) + P (A and B) = 0.88, and that P(neither) = 0.12. We cannot calculate the probability of P(A) unless we know two of the probabilites that add up to 0.88.

Example Question #7 : How To Find The Probability Of An Outcome

a is chosen randomly from the following set: {3, 11, 18, 22} b is chosen randomly from the following set: { 4, 8, 16, 32, 64, 128} What is the probability that a + b = 27?

Since any of the first set can be summed with any of the second set, the addition sign in the equation works like a conjunction. As such, there are 4 * 6 = 24 possible combinations of a and b. Only one of these combinations, 11 + 16 = 27, works. Thus the probability is 1/24, or about 0.04.

Example Question #8 : How To Find The Probability Of An Outcome

There are four aces in a standard deck of playing cards. What is the approximate probability of drawing two consecutive aces from a standard deck of 52 playing cards?

Answer: .005 Explanation: The probability of two consecutive draws without replacement from a deck of cards is calculated as the number of possible successes over the number of possible outcomes, multiplied together for each case. Thus, for the first ace, there is a 4/52 probability and for the second there is a 3/51 probability. The probability of drawing both aces without replacement is thus 4/52*3/51, or approximately .005.

Example Question #2 : Outcomes

In a bag, there are 10 red, 15 green, and 12 blue marbles. If you draw two marbles (without replacing), what is the approximate probability of drawing two different colors?

None of the other answers

Calculate the chance of drawing either 2 reds, two greens, or two blues. Then, subtract this from 1 (100%) to calculate the possibility of drawing a pair of different colors.

The combined probability of RR, GG, and BB is: (10 * 9) / (37 * 36) + (15 * 14) / (37 * 36) + (12 * 11) / (37 * 36)

This simplifies to: (90 + 210 + 132) / 1332 = 432 / 1332

Subtract from 1: 1 - 432 / 1332 = (1332 - 432) / 1332 = approx. 0.6757 or 67.57%

Example Question #10 : How To Find The Probability Of An Outcome

What is the probability of drawing 2 hearts from a standard deck of cards without replacement?

There are 52 cards in a standard deck, 13 of which are hearts

13/52 X 12/51 =

12/ 204 = 3/51 = 1/17

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how to solve probability problems with replacement

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Probability Without Replacement

In these lessons, we will learn how to calculate probability without replacement (dependent events) and how to use a probability tree diagram.

Related Pages Probability Of An Event More Lessons On Probability Probability Tree Diagrams Dependent Events

What Is Probability Without Replacement Or Dependent Probability?

In some experiments, the sample space may change for the different events. For example, a marble may be taken from a bag with 20 marbles and then a second marble is taken without replacing the first marble. The sample space for the second event is then 19 marbles instead of 20 marbles.

This is called probability without replacement or dependent probability. We can use a tree diagram to help us find the probability without replacement.

How To Find The Probability Without Replacement Or Dependent Probability?

Step 1: Draw the Probability Tree Diagram and write the probability of each branch. (Remember that the objects are not replaced) Step 2: Look for all the available paths (or branches) of a particular outcome. Step 3: Multiply along the branches and add vertically to find the probability of the outcome.

Example: A jar consists of 21 sweets. 12 are green and 9 are blue. William picked two sweets at random. a) Draw a tree diagram to represent the experiment.

b) Find the probability that i) both sweets are blue. ii) one sweet is blue and one sweet is green.

c) William randomly took a third sweet. Find the probability that: i) all three sweets are green? ii) at least one of the sweet is blue?

What Is The Difference Between Probability With Replacement (Independent Events) And Probability Without Replacement (Dependent Events) And How To Use A Probability Tree Diagram?

Adam has a bag containing four yellow gumdrops and one red gumdrop. He will eat one of the gumdrops, and a few minutes later, he will eat a second gumdrop. a) Draw the tree diagram for the experiment. b) What is the probability that Adam will eat a yellow gumdrop first and a green gumdrop second? c) What is the probability that Adam will eat two yellow gumdrops? d) What is the probability that Adam will eat two gumdrops with the same color? e) What is the probability that Adam will eat two gumdrops of different colors?

A jar contains 4 black marbles and 3 red marbles. Two marbles are drawn without replacement. a) Draw the tree diagram for the experiment. b) Find probabilities for P(BB), P(BR), P(RB), P(WW), P(at least one Red), P(exactly one red)

Two marbles are drawn without replacement from a jar containing 4 black and 6 white marbles. a) Draw the tree diagram for the experiment. b) Find the probabilities for P(at least one black marble), P(same color), P(BW), P(exactly one black marble)

Probability Question Using Tree Diagrams (Without Replacement)

Example: A bag contains 5 blue balls and 4 red balls. A ball is picked and not replaced. What is the probability of picking at least one red ball?

How To Calculate Probability With And Without Replacement?

A visual tutorial on how to calculate probability with and without replacement using marbles.

How To Calculate Probability Without Replacement Or Dependent Probability?

Example: Andrea has 8 blue socks and 4 red socks in her drawer. She chooses one sock at random and puts it on. She then chooses another sock without looking. Find the probability of the following event P(red, then red).

Mathway Calculator Widget

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Probability Without Replacement
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