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 = $\displaystyle (\frac{8}{10})^4$

2) 3 aces = $\displaystyle _4C_3(\frac{1}{15})^3\frac{14}{15}$

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