Conditional Probability

Conditional probability is a fundamental concept in statistics and probability theory, often used to calculate the likelihood of an event occurring given that another event has already occurred. It helps in understanding complex relationships between different events, especially in fields like data analysis, machine learning, and decision theory. Whether you're studying for exams or applying statistical concepts in real life, conditional probability is essential.

1.0What is Conditional Probability?

Let A and B be the two events associated with the same random experiment. Then, the probability of occurrence of A under the condition B has already occurred and P(B) ≠ 0, is called conditional probability, denoted by P(A | B). 

We define:

$P(A\mid B)=\frac{P(A\cap B)}{P(B)}\text{, where }P(B)>0$

and $P(B\mid A)=\frac{P(B\cap A)}{P(A)}=\frac{P(A\cap B)}{P(A)}\text{, where }P(A)>0$

2.0Conditional Probability Formula

The general formula for conditional probability is:

$P(A\mid B)=\frac{P(A\cap B)}{P(B)}\text{, where }P(B)>0$

This formula assumes that event B has already occurred and helps calculate how likely A is to occur under that condition.

3.0Independent Events and Conditional Probability

If two events, A and B, are Independent Events, then the occurrence of one does not affect the occurrence of the other, i.e.

P(A ⋂ B) = P(A)·P(B)

Then, $\mathrm{P}(\mathrm{A}\mid \mathrm{B})=\frac{\mathrm{P}(\mathrm{A})\cdot \mathrm{P}(\mathrm{B})}{\mathrm{P}(\mathrm{B})}=\mathrm{P}(\mathrm{A})$

$P(B\mid A)=\frac{P(A)\cdot P(B)}{P(A)}=P(B)$

If A and B are Dependent Event, then

$\mathrm{P}(\mathrm{A}\cap \mathrm{B})=\mathrm{P}(\mathrm{A})\mathrm{P}\left(\frac{\mathrm{B}}{\mathrm{A}}\right)$

$P(A\cap B)=P(B)P\left(\frac{A}{B}\right)$

Example: Tossing an unbiased coin and rolling a die are independent events. If we ask, "What is the probability of getting a head on the coin, given that we rolled a 4?" Since these events are independent, the probability remains:

$P(\text{ Heads }\mid \text{ Roll }4)=P(\text{ Heads })=\frac{1}{2}$

4.0Joint Probability and Conditional Probability

The joint probability P(A ⋂ B) represents the probability that both events A and B occur. The conditional probability formula can also be derived from the joint probability: P(A ⋂ B) = P(A|B) × P(B)

This relationship is essential when dealing with multiple dependent events.

5.0Bayes' Theorem and Conditional Probability

Bayes' theorem calculates the probability of an event occurring given specific conditions or prior information. It is used to determine conditional probabilities and is often referred to as the formula for assessing the likelihood of causes. 

The formula for Bayes' theorem is:

$P(A\mid B)=\frac{P(B\mid A)\cdot P(A)}{P(B)}$

6.0Solved Examples on Conditional Probability

Example 1: if an odd number comes up on tossing a die. Find the probability of its being a prime number.

Solution: 

S = {1, 2, 3, 4, 5, 6}

A = {1, 3, 5}

B = {2, 3, 5}

Now $\mathrm{P}(\mathrm{B}\mid \mathrm{A})=\frac{\mathrm{P}(\mathrm{A}\cap \mathrm{B})}{\mathrm{P}(\mathrm{A})}$

We know that A ⋂ B = {3, 5}

P(A ⋂ B) =$\frac{2}{6}=\frac{1}{3}$

$\mathrm{P}(\mathrm{A})=\frac{3}{6}=\frac{1}{2}$

Then, $P(B\mid A)=\frac{2}{3}$

Example 2: A box contains 4 bad and 6 good tubes. Two are drawn out from the box at a time. One of them is tested and found to be good. What is the Probability that the other one is also good?

Solution: 

$P\left(\frac{G}{G}\right)=\frac{P(G\cap G)}{P(G)}$

$P\left(\frac{G}{G}\right)=\frac{\frac{{}^6{C}_2}{{}^{10}{C}_2}}{\frac{{}^6{C}_1}{{}^{10}{C}_1}}$

$P\left(\frac{G}{G}\right)=\frac{6\times 5\times 10}{6\times 10\times 9}$

$P\left(\frac{G}{G}\right)=\frac{5}{9}$

Example 3: In a certain school, 20% of the students failed in English, 15 % of the student failed in Mathematics and 10% of the student failed both in English and Mathematics. A student is selected at random. If he failed in Mathematics. What is the probability that he also failed in English?

Solution:

P(E)= 0.2

P(M) = 0.5

P(E ⋂ M) = 0.1

$P\left(\frac{E}{M}\right)=\frac{P(E\cap M)}{P(M)}$

$P\left(\frac{E}{M}\right)=\frac{0.1}{0.15}$

$P\left(\frac{E}{M}\right)=\frac{10}{15}$

$P\left(\frac{E}{M}\right)=\frac{2}{3}$

Example 4: An urn contains 10 white and 5 black balls. Two balls are drawn sequentially from an urn without replacement. What is the probability that both balls drawn are white?

Solution: 

$\mathrm{P}(\mathrm{W}\cap \mathrm{W})=\mathrm{P}(\mathrm{W})\mathrm{P}\left(\frac{\mathrm{W}}{\mathrm{W}}\right)$

$\mathrm{P}(\mathrm{W}\cap \mathrm{W})=\frac{{}^{10}{C}_1}{{}^{15}{C}_1}\cdot \frac{{}^9{C}_1}{{}^{14}{C}_1}$

$\mathrm{P}(\mathrm{W}\cap \mathrm{W})=\frac{10\times 9}{15\times 14}$

$P(W\cap W)=\frac{3}{7}$

Example 5: Two cards are drawn one by one without replacement from a well- shuffled pack of 52 cards. What is the probability that one is the red queen and the other is a king of black.

Solution: 

$P(P\cap K)=P(Q)P\left(\frac{K}{Q}\right)+P(K)P\left(\frac{Q}{K}\right)$

$\mathrm{P}(\mathrm{P}\cap \mathrm{K})=\frac{{}^2{C}_1}{{}^{52}{C}_1}\cdot \frac{{}^2{C}_1}{{}^{51}{C}_1}+\frac{{}^2{C}_1}{{}^{52}{C}_1}\cdot \frac{{}^2{C}_1}{{}^{51}{C}_1}$

$\mathrm{P}(\mathrm{P}\cap \mathrm{K})=\frac{4}{52\times 51}+\frac{4}{52\times 51}$

$P(P\cap K)=\frac{8}{52\times 51}$

$\mathrm{P}(\mathrm{P}\cap \mathrm{K})=\frac{2}{663}$

7.0Practice Questions on Conditional Probability

1. Two dice were thrown, and it is known that the numbers which came up were different. Find the probability that the sum of the two numbers was 4.
2. Ten cards, each numbered 1 to 10 are placed in a box, thoroughly shuffled, and one card is drawn at random. Given that the number on the drawn card is greater than 3, what is the probability that the card shows an even number?
3. Assuming each child has an equal likelihood of being a boy or a girl, consider a family with two children. What is the conditional probability that both children are girls given that (i) the younger child is a girl, and (ii) at least one of the children is a girl?
4. An instructor has a question bank containing 30 easy true/false questions, 20 difficult true/false questions, 50 easy multiple-choice questions, and 40 difficult multiple-choice questions. If a question is randomly selected from the bank, what is the probability that it is an easy question, given that it is a multiple-choice question?
5. Three coins are tossed simultaneously. Find the probability that all coins show heads if at least one of the coins shows a head.

8.0Sample Questions on Conditional Probability

1. What is the formula for Conditional Probability?

Ans: The formula for conditional probability is:

$P(A\mid B)=\frac{P(A\cap B)}{P(B)}$

Where:

• P(A ⋂ B) represents the probability that both events A and B happening.
• P(B) denotes the probability that event B occurs, with P(B) > 0.

2. What is Bayes' Theorem?

Ans: Bayes' theorem is a way to reverse conditional probabilities. It helps in updating probabilities based on new information. The formula for Bayes’ theorem is:

$P(A\mid B)=\frac{P(B\mid A)\cdot P(A)}{P(B)}$

This theorem is widely used in statistics, machine learning, and decision theory to make inferences from data.

3. How do you calculate conditional probability for dependent events? 

Ans: To calculate conditional probability for dependent events, use the formula:

$P(A\mid B)=\frac{P(A\cap B)}{P(B)}$

This formula adjusts the probability of A based on the known occurrence of event B.