A drawer holds 4 red hats and 4 blue hats. What is the probability of getting…

Concept: for n independent trials each with success probability p, the probability of exactly k successes is the binomial probability C(n,k)·p^k·(1−p)^(n−k). Drawing a hat and immediately replacing it before the next draw keeps every draw independent and identical, so this formula applies directly.

1. Since the hat is replaced every time, the drawer always holds 4 red and 4 blue hats out of 8, so the probability of red on any single draw is 1/2 and the probability of blue is 1/2, across all 4 draws.

2. Probability of exactly 3 red hats and 1 blue hat in 4 draws = C(4,3)·(1/2)^3·(1/2)^1 = 4·1/16 = 1/4.

3. By the same formula with red and blue swapped, the probability of exactly 3 blue hats and 1 red hat = C(4,3)·(1/2)^1·(1/2)^3 = 4·1/16 = 1/4.

4. Getting exactly 3 red hats and getting exactly 3 blue hats cannot both happen in the same 4 draws, so these are mutually exclusive events; add their probabilities: 1/4 + 1/4 = 1/2.

Cross-check: because each draw has probability 1/2 for either colour, the binomial distribution over the number of red hats in 4 draws is symmetric, so P(exactly 3 red) = P(exactly 1 red) = 4/16, and P(exactly 3 blue) = P(exactly 1 red) too, by the same symmetry. This independent route again gives 4/16 + 4/16 = 8/16 = 1/2, confirming the result.

Answer: 1/2.