# Math Help - archer probability

1. ## archer probability

An archer hits the bull’s eye 80% of the time. If he shoots 5 arrows, find the probability that he will get 4 bull’s eyes

to answer this question would you just multiply

his success rate? ie

.8 * .8 * .8 * .8 = 0.4096 so about 40 % chance with five
arrows ?

2. You have to multiply the probability of 4 hits and 1 miss (0.8^4(1-0.8)) with the number of ways this can happen in:

Miss - Hit - Hit - Hit - Hit
Hit - Miss - Hit - Hit - Hit
Hit - Hit - Miss - Hit - Hit
Hit - Hit - Hit - Miss - Hit
Hit - Hit - Hit - Hit - Miss

p(4 hits)=0.8^4(1-0.8)*5=40,96%

So your answer is right (because 0.2*5=1), but try with another hit rate.

3. Originally Posted by opsahl92
You have to multiply the probability of 4 hits and 1 miss (0.8^4(1-0.8)) with the number of ways this can happen in:

Miss - Hit - Hit - Hit - Hit
Hit - Miss - Hit - Hit - Hit
Hit - Hit - Miss - Hit - Hit
Hit - Hit - Hit - Miss - Hit
Hit - Hit - Hit - Hit - Miss

p(4 hits)=0.8^4(1-0.8)*5=40,96%

So your answer is right (because 0.2*5=1), but try with another hit rate.
The calculation you give does not have that answer, but the answer is correct. Each of the 5 ways of getter 4 hits has the same probability and so:

$p(4\text{ hits})= 5 \times 0.8^4 \times 0.2=0.4096$

or $40.96\%$

Equally you could observe that the probability of the number of hits has a binomial distribution $B(5,0.8)$, and so:

$p(n\text{ hits})=b(n;5.0.8)$

which will give the same result when $n=4$.

CB

4. To avoid confusion for the thread starter, I just have to point out that there is nothing wrong with my calculation. I'll make it LaTeX though, for easy reading:

$p(\text{4 hits})=0.8^4\times(1-0.8)\times5=0,4096$