Question 1194069
This would be 0.82^4*0.18=0.0814
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for the first shot it would be 0.18 probability to score
for the second shot it would be 0.82*0.18=0.1476, so the answer is the sum or 0.3276 probability
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the expected number of shots before scoring is 1/p=1/0.18, or 5.56