Questions: Conditional Probability

Question 1

A bag contains 3 red balls and 2 blue balls. You draw one ball and observe it is red. Without replacing it, what is the probability the next draw is also red?

• A. 3/5
• B. 1/2
• C. 2/5
• D. 3/10

Answer: (B)

After seeing the first ball is red, the sample space shrinks to all outcomes where that happened. Now 4 balls remain (2 red, 2 blue), so P(2nd red | 1st red) = 2/4 = 1/2. Using the formula: P(both red) = (3/5)(2/4) = 6/20, P(1st red) = 3/5 = 12/20, so P(A|B) = (6/20)/(12/20) = 1/2.

Question 2

For any two events A and B with P(A) > 0 and P(B) > 0, it is typically true that P(A|B) = P(B|A).

• T. True
• F. False

Answer: False

P(A|B) = P(A∩B)/P(B) and P(B|A) = P(A∩B)/P(A). These are equal only when P(A) = P(B). Consider: P(ground is wet | it is raining) is very high, but P(it is raining | ground is wet) is lower because sprinklers or other causes can also wet the ground. This asymmetry is the core insight behind Bayes' theorem.

Question 3

Explain in your own words why the formula P(A|B) = P(A∩B)/P(B) divides by P(B).

Model answer: Dividing by P(B) rescales the probabilities of the remaining outcomes (those inside B) so they sum to 1, forming a valid probability distribution on the restricted sample space.

When we condition on B, we discard all outcomes where B did not occur. The outcomes in A∩B now represent the only relevant cases, but their probabilities originally summed to P(B), not 1. Dividing by P(B) rescales them to form a proper probability distribution. Without this rescaling, the conditional probabilities would not sum to 1 and would violate the probability axioms.