Originally Posted by

**jacs** A marksman finds that on average he hits the target 9 times out of every 10 and scores a bull's eye on the average once every 5 rounds. He fires 4 rounds. What is the probability that:

He scores at least 2 bull's eyes and he has hit the target on each of the four rounds?

I have found that the prob he scores at least 2 bulls eyes to be 0.1808

and the prob of hittign the target all four times to be 0.6561

what i am not sure how to do is combine them

Probability of a b-eye given that the target is hit is $\displaystyle 2/9$,

so the probability of 2 b-eyes in 4 hits is:

$\displaystyle

P(2 \mbox{b-eyes from 4 hits})={4 \choose 2} (2/9)^2 (7/9)^2 \approx 0.1792

$

The probability of four hits from four shots is $\displaystyle 0.9^4 \approx 0.6561$.

So the probability of 2 b-eyes and four hits from four shots is:

$\displaystyle

P(2 \mbox{b-eyes and 4 hits from 4 shots})=$$\displaystyle P(\mbox{2 b-eyes from 4 hits})$$\displaystyle P(\mbox{4 hits from 4 shots})$

$\displaystyle

\approx 0.1792 \times 0.6561 \approx 0.1176

$

RonL