Question 1143458
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P(0 hits) = C(3,0)*(.368^0)*(1-.368)^3
P(1 hit)  = C(3,1)*(.368^1)*(1-.368)^2
P(2 hits) = C(3,2)*(.368)^2*(1-.368)^1
P(3 hits) = C(3,3)*(.368)^3*(1-.368)^0<br>
The answers to parts A and B are in the above calculations.<br>
For part C, you can either add the probabilities for 1, 2, and 3 hits, or you can just do 1 minus the probability of 0 hits.