Answer:
[tex]P(X=2) = 0.1047[/tex]
Step-by-step explanation:
Represent the probability of hitting the target with H.
So:
[tex]H = \frac{1}{8}[/tex]
Represent the probability of not hitting the target with T.
So:
[tex]T = 1 - H[/tex]
[tex]T =1 - \frac{1}{8}[/tex]
[tex]T =\frac{7}{8}[/tex]
Required
Probability of exactly 2 hits in 5 tries
This probability is a binomial probability and will be calculated using:
[tex]P(X=x) = ^nC_x H^xT^{n-x}[/tex]
In this case:
[tex]x = 2[/tex] i.e 2 hits
[tex]n = 5[/tex] i.e. 5 attempts
So, we have:
[tex]P(X=2) = ^5C_2 H^2T^{5-2}[/tex]
[tex]P(X=2) = ^5C_2 H^2T^3[/tex]
Apply combination formula:
[tex]P(X=2) = \frac{5!}{(5-2)!2!} H^2T^3[/tex]
[tex]P(X=2) = \frac{5!}{3!2!} H^2T^3[/tex]
[tex]P(X=2) = \frac{5*4*3!}{3!2*1} H^2T^3[/tex]
[tex]P(X=2) = \frac{5*4}{2} H^2T^3[/tex]
[tex]P(X=2) = 10* H^2T^3[/tex]
Substitute values for H and T
[tex]P(X=2) = 10* (\frac{1}{8})^2(\frac{7}{8})^3[/tex]
[tex]P(X=2) = 10* \frac{1}{64}*\frac{343}{512}[/tex]
[tex]P(X=2) = \frac{3430}{32768}[/tex]
[tex]P(X=2) = 0.1047[/tex]
Hence, the probability of 2 hits in 5 tries is 0.1047
