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- Probability with replacement – Explanation & Examples

## Probability with replacement and independence:

After reading this article, you should be able to:

- Understand what probability with replacement means.
- Difference between probability with and without replacement.
- Calculate probability with replacement using basic probability theory.
- Calculate probability with replacement using tree diagrams.

## What does probability with replacement mean

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

- Draw a ball; it could be blue or orange.
- Replace the ball from the first draw. So after replacement, the box again contains 3 orange and 2 blue balls.
- Draw another ball that could again be either blue or orange.

Method 2 (Without replacement):

- Draw a ball, and it could be blue or orange.
- Do not replace the ball from the first draw and draw another ball that could again be either blue or orange.

## How to calculate Probability with replacement:

Event1 = First ball is orange, and

Event2 = Second ball is orange

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

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

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

$ 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

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

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

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

- Both Balls are Green.
- Both balls are Red.
- The first ball is Green and the second is Red.
- The first ball is Red and the second is Green.

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

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

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

All four samples are independent, so

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

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

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

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

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

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

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

Event1: The first book is not Maths

Event2: Second book is not Maths

Event3: Third book is not Maths

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

2. C’ represents Not the letter C.

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

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

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## Probability with Replacement

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

(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.

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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

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## Probability with replacement marbles

## 3 Answers 3

Yes, you are on a right track:

Total number of balls always remains $9$ .

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

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## GRE Math : Outcomes

The probability of drawing two red marbles

The probability of drawing exactly one blue marble.

The relationship cannot be determined from the information given.

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

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

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

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

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?

Probability of drawing green from A:

Probability of drawing green from B:

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

Quantity A: The probability that event A will occur.

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

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

## Example Question #2 : Outcomes

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

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## Probability Without Replacement

## What Is Probability Without Replacement Or Dependent Probability?

## How To Find The Probability Without Replacement Or Dependent Probability?

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

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

## Probability Question Using Tree Diagrams (Without Replacement)

## How To Calculate Probability With And Without Replacement?

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

## IMAGES

## VIDEO

## COMMENTS

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