Four fair dice are thrown independently 27 times. Then the expected number of times, at least two dice show up a three or a five, is _________.

Solution

Maths sir · 4 steps

Find the chance that one throw of four dice is a success, then multiply by 27 throws.

1. One die. A die shows a three or a five in 2 of its 6 faces, so p = 1/3 and q = 2/3. The number of such dice among the four is binomial with n = 4.
2. Use the complement. At least two means not 0 and not 1. P(0) = (2/3)⁴ = 16/81 and P(1) = 4 × 1/3 × (2/3)³ = 32/81.

   P(at least two) = 1 - 16/81 - 32/81 = 33/81 = 11/27

3. Expected count. Each of the 27 throws succeeds with probability 11/27, so the number of successes is binomial with mean np = 27 × 11/27.
4. So. The expected number of times is 11.

The common slip: 19. Forgetting the factor C(4, 1) = 4 in P(1) gives 57/81 for one throw and 19 for the expected count, not 11.

The idea to keep. When trials are repeated, first find the success probability of one trial, then the expected count is just np. For at least two, subtract the 0 and 1 cases from 1.

Answer from the NTA final answer key. Solution by PrepWiser.