Answer to Question #123804 in Statistics and Probability for Yasini

Question

A. Suppose that a marksman hits the bull's eye 15000 times out of 50000 shots. If the next 4 shots are independent. Find the probability that:
(I)the next 4 hit the bull's eye
(Ii) two of the next shots hits the bull's eye

Answer 1

The probability that a marksman hits the bull's eye is:

$p=\frac{15000}{50000}=0.3$

(i) We are dealing with 4 independent events. Using the rule of multiplication, we find that the probability is $p^4=(0.3)^4=0.0081$

(ii) Since two successful shots may be at arbitrary places (e.g., first and second shots can be successful, or second and fourth, etc) we have to multiply by the binomial coefficient $C_4^2=\frac{4!}{2!2!}$ .(a number of possible subsets of two elements that are chosen from the set of four elements) The resulting probability is $C_4^2p^2(1-p)^2=\frac{4!}{2!2!}(0.3)^2(0.7)^2=0.2646$

Answer: (i) $0.0081$ (ii) $0.2646$