Is it easier to flip three coins and get exactly two heads or is it easier to flip four coins and get exactly two heads? Explain

Answer 1

Let's take the three coins case first:

In order to get exactly two heads in 3 flips, the probability is 3/8. This is because there are 3 possibilities (HHT, HTH, THH) and each of those possibilities occurs with 1/8 (1/2 * 1/2 * 1/2) probability.

In order to get exactly two heads in four flips, the probability is 6/16 or 3/8. This is because there are 6 possibilities (HHTT, HTHT, HTTH, THHT, THTH, TTHH) and each one occurs with 1/16 (1/2 * 1/2 * 1/2 * 1/2) probability.

So the chance of getting 2 heads with three flips or with four flips is the same.

Note that this can also be solved with the binomial theorem. The theorem states that the probability of getting k of some outcome that has a probability of p in n trials is (n choose k)(p)^k(1-p)^(1-k). So in this case,

3 flips: (3 choose 2)(1/2)²(1/2)¹ = 3 * 1/8 = 3/8

4 flips: (4 choose 2)(1/2)²(1/2)² = 6 * 1/16 = 3/8