# Thread: [SOLVED] Binomial Probability with compound events

1. ## [SOLVED] Binomial Probability with compound events

A tennis player finds that on average he gets his serve in eight out of every ten attempts. He serves an ace once every fifteen serves. He serves four times.
Assuming that successive serves are independent events find the probability that he hits exactly three aces and the other serve lands in.

My approach was:
1) All serves in = $(\frac{8}{10})^4$
2) 3 aces = $_4C_3(\frac{1}{15})^3\frac{14}{15}$

Multiply the 2 results = 0.000453

However the solutions have 0.000869
May I ask where is it that I'm going wrong

TIA for any feedback

2. First find probability of getting a serve in that's not an ace. To make things concrete, take 30 serves. Then 24 go in (8 out of 10), and 2 are aces (1 in 15). That means 22 of 30 serves go in, but are not aces, i.e., probability = 11/15.

The probability you want is 4C3 * prob of 3 aces * prob of in but not ace =
4 * (1/15)^3 * (11/15) = 0.000869