Answer to Question #192368 in Statistics and Probability for jordan

Question

In a shooting competition, Mr X can shoot at the bulls eye 4 times out of 5 shots and

Mr Y, 5 times out of six and Mr Z, 3 times out of 4 shots. Find the probability that

the target will be hit at least twice?

Answer 1

Probability of X hitting the target P(X) $=\dfrac{4}{5}=0.8$

$P(X')=1-P(X)=1-0.8=0.2$

Probability of Y hitting the target P(Y) $=\dfrac{5}{6}=0.83$

$P(Y')=1-P(Y)=1-0.83=0.17$

Probability of Z hitting the target $P(Z) =\dfrac{3}{4}=0.75$

$P(Z')=1-P(Z)=1-0.75=0.25$

Let 'H' be the random and denoting the number of target hit

Probability that the target will be hit at least twice-

$P( H\ge 2) =1-P(H<2)=1-[P(H=0)+P(H=1)]$

$=1- [P(X')P(Y')P(Z')+P(X')P(Y')P(Z)+P(X')P(Y)P(Z')+P(X)P(Y')P(Z')]$

$=1-[(0.2)(0.17)(0.25)+(0.2)(0.17)(0.75)+(0.2)(0.83)(0.25)+(0.8)(0.17)(0.25)] \\[9pt]=1-[0.0085+0.0255+0.0415+0.034]\\[9pt]=1-0.1095=0.8905$