How to Find Probability without Replacement

The term "without replacement" in probability describes a situation in which every item taken out of a set is not returned to the set before the next draw.

There are different real-life applications of this concept such as card games, sampling, and resource allocation. For drawing k items from a set of n items without replacement, the probability of a specific sequence of outcomes can be calculated as

P = Number of Favorable Sequences/ Total Number of Possible Sequences

In this article, we will discuss how to find probability without replacements, its real-life application, and some practice problems on it.

Fundamentals of Probability

To understand how to calculate probability without replacement, it's essential to grasp some fundamental definitions and concepts:

• Probability: The measure of the likelihood that an event will occur. It is calculated as the ratio of favourable outcomes to the total number of possible outcomes.

• Event: An outcome or a set of outcomes from a random experiment.

• Sample Space: The set of all possible outcomes in a probability experiment.

• Without Replacement: A scenario where items are not returned to the pool after being selected, affecting subsequent selections.

Differences Between Replacement and Non-Replacement Scenarios

The sample space changes with each selection, and this is the main distinction between replacement and non-replacement scenarios:

With Replacement: Every choice stands alone from the ones that came before it. For every selection, the sample space stays the same.

Without Replacement: Each decision influences the ones that follow. Each selection results in a reduction of the sample space, which makes the events reliant.

How to Find Probability Without Replacement

To finding probability without replacement follow the steps added below:

Step 1: Define Sample Space: Prior to making any decisions, ascertain the entire number of possible outcomes.

Step 2: Find Favorable Outcomes: Ascertain how many outcomes the event of interest has that are favorable.

Step 3: Calculate the Probability: Apply the formula P(A) = Number of Favorable Outcomes / Total Number of Possible Outcomes

Step 4: Adjust for Subsequent Selections: Update the sample space following each selection, then repeat the computation for any more occurrences.

Examples on Probability without Replacement

Example 1: Drawing Balls from a Bag

Let's say we have three blue and five red balls in a bag. Two balls are drawn without replacement. What is the probability of drawing a blue ball and then a red ball?

Solution:

First Draw: Probability of drawing a red ball:

P(Red) = 5/8

Second Draw: After drawing a red ball, 7 balls remain (4 red and 3 blue). Probability of drawing a blue ball:

P(Blue | Red) = 3/7

Combined probability:

P(Red and Blue) = P(Red) × P (Blue | Red) = 5 /8 × 3 / 7 = 15 / 56

Example 2: Selecting Cards from a Deck

A standard deck of 52 cards is shuffled. What is the probability of drawing an Ace followed by a King without replacement?

Solution:

First Draw: Probability of drawing an Ace:

P (Ace) = 4 / 52 = 1 / 13

Second Draw: After drawing an Ace, 51 cards remain. Probability of drawing a King:

P( King | Ace) = 4 / 51

Combined Probability:

P(Ace and King) = P(Ace) × P(King | Ace) = 1 / 13 × 4 / 51 = 4 / 663

Conditional Probability in Non-Replacement Scenarios

The likelihood of an event happening in the presence of another event that has already happened is known as Conditional probability.

Conditional probability is crucial in non-replacement settings since every choice has an impact on the ones that come after.

Example: Conditional Probability with Marbles

We have a bag containing 6 green marbles and 4 yellow marbles. We draw two marbles without replacement. What is the probability of drawing two green marbles?

Solution:

First Draw: Probability of drawing a green marble:

P(Green₁) = 6 / 10 = 3/5

Second Draw: After drawing a green marble, 9 marbles remain (5 green and 4 yellow). Probability of drawing another green marble:

P(Green₂ | Green₁) = 5/9

Combined Probability:

P(Green₁ and Green₂) = P(Green₁) × P(Green₂ | Green₁) = 3 / 5 × 5/9 = 1/3

Problems on Probability without Replacement

Problem 1: In a bag of 5 red, 4 yellow, and 3 green marbles, what is the probability of drawing a yellow marble followed by a green marble without replacement?

Solution:

Probability of drawing the first yellow marble:

P(Yellow) = 4/12 = 1/3

Probability of drawing a green marble given a yellow marble was drawn:

P(Green | Yellow) = 3/11

Combined Probability:

P(Yellow and Green) = 1/3 × 3/11 = 1/11

Problem 2: A deck of cards is shuffled. What is the probability of drawing a spade followed by a heart without replacement?

Solution:

Probability of drawing a spade:

P(Spade) = 13/52 = 1/4

Probability of drawing a heart given a spade was drawn:

P(Heart | Spade) = 13/51

Combined Probability:

P(Spade and Heart) = 1/4 × 13/51 = 13/204

Problem 3: A jar contains 9 blue, 7 red, and 4 green balls. Two balls are drawn without replacement. What is the probability of drawing a red ball followed by a green ball?

Solution:

Probability of drawing the first red ball:

P(Red) = 7/20

Probability of drawing a heart given a spade was drawn:

P(Green | Red) = 4/19

Combined Probability:

P(Red and Green) = 7/20 × 4/19 = 28/380 = 7/95

Practice Questions on Probability without Replacement

Q1. A jar contains 6 green and 5 blue marbles. Two marbles are drawn without replacement. What is the probability of drawing a green marble followed by a blue marble?

Answer: 3/11

Q2. In a jar of 8 black and 6 white marbles, what is the probability of drawing a black marble followed by another black marble without replacement?

Answer: 4/13

Q3. In a deck of 52 cards, what is the probability of drawing a diamond followed by a spade without replacement?

Answer: 13/204

Q4. A jar contains 5 yellow, 4 green, and 3 blue marbles. Two marbles are drawn without replacement. What is the probability of drawing a green marble followed by a yellow marble?

Answer: 5/33

Q5. A box contains 8 white and 6 black pens. What is the probability of drawing a black pen followed by a white pen without replacement?

Answer: 24/61

Challenges on Finding Probability

various challenges on finding probability includes:

Misunderstanding Problem

One frequent difficulty is misinterpreting the issue, particularly when attempting to distinguish between scenarios involving replacement and those without. Make sure to state the problem clearly and indicate if the choices are independent or dependent.

Incorrect Calculations

Frequently, inaccurate computations result from failing to update the sample space following every selection. Remind yourself to change the total number of items following each draw in order to guarantee precise probability computations.

Also Read:

• How to Calculate Probability
• Conditional Probability
• Tricks to solve Probability Question

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Article Tags :

• Mathematics
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• Probability