Question

suppose that a marksman hits the bull's eye 15,000 times outshots.If the next 4 shots are independent.Find the probability that

1.the next 4 shot hit the bull's eye.

2.Two of the next shot hits the bull's eye

Answer 1

We assume that a marksman made n shots and 15000 of them were successful. Thus, a probability to make a successful outshot is $p=\frac{15000}{n}.$ Since we have four independent shots, we may use a binomial distribution with parameters n=4 and $p=\frac{15000}{n}$ . Then, we receive:

1. The probability that 4 outshots are successful is $(\frac{15000}{n})^4$ .
2. The probability that 2 of 4 shots are sucessful is $C_4^2(\frac{15000}{n})^2(1-\frac{15000}{n})^2=6(\frac{15000}{n})^2(1-\frac{15000}{n})^2$ , where $C_4^2=\frac{4!}{2!2!}=6$ is a binomial coefficent.

Answer:1. $(\frac{15000}{n})^4$ 2. $6(\frac{15000}{n})^2(1-\frac{15000}{n})^2$ ; $n$ is a number of shots that a marksman made ( not including the last 4 shots).