Question
The probability of a man hitting a target is 1 5 . If the man fires 7 times, then what is the probability that he hits the target at least twice?
The probability of a man hitting a target is 1 5 . If the man fires 7 times, then what is the probability that he hits the target at least twice?
C. 1 - ( 11 5 ) ( 4 5 )^6
Let p be the probability of hitting the target and q be the probability of missing the target. p = 1 5 q = 1 - p = 4 5 Number of trials, n = 7 Let X be the random variable representing the number of hits. The probability of hitting the target at least twice is given by: P(X 2) = 1 - P(X P(X 2) = 1 - [P(X = 0) + P(X = 1)] Using the binomial distribution formula P(X = r) = ^ n C_ r p^r q^ n-r : P(X = 0) = ^ 7 C_ 0 ( 1 5 )^0 ( 4 5 )^7 = ( 4 5 )^7 P(X = 1) = ^ 7 C_ 1 ( 1 5 )^1 ( 4 5 )^6 = 7 5 ( 4 5 )^6 Substituting these values: P(X 2) = 1 - [ ( 4 5 )^7 + 7 5 ( 4 5 )^6 ] P(X 2) = 1 - ( 4 5 )^6 [ 4 5 + 7 5 ] P(X 2) = 1 - ( 11 5 ) ( 4 5 )^6 Answer: 1 - ( 11 5 ) ( 4 5 )^6
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