mug - 8 months ago 46

R Question

Harry is a basketball player.His probability of making a free throw is 0.70.

-> what is the probability that Harry makes his third free throw on his fifth shot?

things I have tried so far; given p=0.70 n = 5

`sum(dbinom(3:5,n,p))`

And I got 0.83, but I know that answer is 0.18.

`pbinom(5,n,p)-pbinom(2,n,p)`

Can someone please help me.

Answer

If he gets 3rd free throw on 5th shot, then

- the previous 4 shots have 2 free throws;
- the 5th shot is a free throw

Note, these two events are independent, while the probability of the first is `dbinom(2, 4, 0.7)`

, and the probability of the last is 0.7. So

```
dbinom(2, 4, 0.7) * 0.7
# [1] 0.18522
```

Note, there is really a trap here. It is so easy to assume the answer as

```
dbinom(3, 5, 0.7)
# [1] 0.3087
```

But that only means we have 3 free throws out of 5 shots, which does not guarantees the 5th shot is a free throw.

Source (Stackoverflow)