Question

The probability that a man hits a target is 1/3.



a) if he fires five times, what is the probability of his hitting target at least two times?



b) how many times must he fire so that the probability of his hitting target at least once is more than 90%?

Answer 1

Hope this help..........

Answer 2

Answer:

The probability that a man hits a target is 1/3.

a) if he fires five times, what is the probability of his hitting target at least two times?

We will use binomial

Formula : $P(X=r)=^nC_rp^rq^{n-r}$

Probability of success = p = 1/3

Probability of failure = 1-1/3 = 2/3

Probability of  his hitting target at least two times :

= 1-[ P(X=0)+P(X=1)}

=$1-[(^5C_0 (\frac{1}{3})^0 (\frac{2}{3})^5)+^5C_1 (\frac{1}{3})^1  (\frac{2}{3})^4)$

=$1-[(\frac{5!}{0!(5-0)!} (\frac{1}{3})^0 (\frac{2}{3})^5)+\frac{5!}{1!(5-1)!} (\frac{1}{3})^1  (\frac{2}{3})^4)$

=$0.539$

b) how many times must he fire so that the probability of his hitting target at least once is more than 90%?

$P(x\geq 1)>0.9$

$1-P(x= 0)>0.9$

$1-(^rC_0 (\frac{1}{3})^0 (\frac{2}{3})^5)>0.9$

$1-(\frac{r!}{0!(r-0)!} (\frac{1}{3})^0 (\frac{2}{3})^r)>0.9$

r>5.6

So, he must fire more than 5 times so that the probability of his hitting target at least once is more than 90%