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?

Step-by-step solution

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) = ⁷C₀ ( 1 5 )^0 ( 4 5 )^7 = ( 4 5 )^7 P(X = 1) = ⁷C₁ ( 1 5 )^1 ( 4 5 )^6 = 7 5 ( 4 5 )^6 Substituting these values: