Hello, I have been working on this problem for a while and just can not seem to

Question

Hello,

I have been working on this problem for a while and just can not seem to get the last one part (d). I have included my answers for the other parts if they help any, Thanks ahead of time for your help!

Question: Two buddies, Zeke and Bubba, plan a squirrel-hunting trip. Bubba is a better shot than Zeke. How much better? Bubba is 2-times a better shot than Zeke. Upon seeing a squirrel, Bubba and Zeke raise their rifles and simultaneously shoot at the squirrel. Assume that Bubba shooting the squirrel and Zeke shooting the squirrel are independent events, and that the probability of Zeke shooting the squirrel is 0.36.

Part (a) What is the probability that the squirrel will be shot? 0.8208

Part (b) What is the probability that both Zeke and Bubba will miss the squirrel? 0.1792

Part (c) If the squirrel is shot, what is the probability that Zeke shot the squirrel? 0.4386

Part (d) If only one of the two buddies shot the squirrel, what is the probability it was Bubba?

Explanation / Answer

Z: Zeke shooting the squirrel, P(Z)=0.36

B: Bubba shooting the squirrel. It is given that P(B)=2*P(Z)=0.72

a) P(Squirrel will be shot) =1-P(both didn't shoot the squirrel)=1-(1-0.36)*(1-0.72)=0.8208

b) P(both didn't shoot)=(1-0.36)*(1-0.72)= 0.1792

c) P(Z|squirrel is shot)=P(Z shot the squirrel)/P(Squirrel will be shot)=0.36/0.8208=0.4386

d) P(only one shoots)=P(B)*(1-P(Z))+P(1-B)*P(Z) =(0.36)*(1-0.72)+(1-0.36)*(0.72) =0.5616

P(B shoots and Z doesn't)/P(only one shoots) =(0.36)*(1-0.72)/0.5616 =0.1795

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