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What is the probability that 2 or more of these t-shirts are irregular?

A t-shirt manufacturing company advertises to its retailers that the probability of an individual t-shirt being first quality is 95%. A box of 15 such t-shirts is selected for inspection. What is the probability that 2 or more of these t-shirts are irregular? I need some help with this problem.

Public Comments

1. This is solved with the binomisl formula. First find the probability of finding x = 0 defective t-shirts. P(x = 0) = n!/(0!(n-0)!)p^0(1-p)^(n-0) = 0.05^0(0.95)^15 = 0.4633 and P(x = 1)= 15!/14!0.05^1(0.95)^14 = 0.3657 So the probability of 2 or more irregulars-- P(x ≥ 2) = 1 - [0.4633 + 0.3657] = 0.171
2. Let X be the number of t-shirts that are irregular. X has the binomial distribution with n = 15 trials and success probability p = 0.05 In general, if X has the binomial distribution with n trials and a success probability of p then P[X = x] = n!/(x!(n-x)!) * p^x * (1-p)^(n-x) for values of x = 0, 1, 2, ..., n P[X = x] = 0 for any other value of x. The probability mass function is derived by looking at the number of combination of x objects chosen from n objects and then a total of x success and n - x failures. Or, in other words, the binomial is the sum of n independent and identically distributed Bernoulli trials. X ~ Binomial( n , p ) the mean of the binomial distribution is n * p = 0.75 the variance of the binomial distribution is n * p * (1 - p) = 0.7125 the standard deviation is the square root of the variance = √ ( n * p * (1 - p)) = 0.8440972 The Probability Mass Function, PMF, f(X) = P(X = x) is: P( X = 0 ) = 0.4632912 P( X = 1 ) = 0.3657562 P( X = 2 ) = 0.1347523 P( X = 3 ) = 0.03073298 P( X = 4 ) = 0.004852576 P( X = 5 ) = 0.0005618772 P( X = 6 ) = 4.928747e-05 P( X = 7 ) = 3.335243e-06 P( X = 8 ) = 1.755391e-07 P( X = 9 ) = 7.18581e-09 P( X = 10 ) = 2.269203e-10 P( X = 11 ) = 5.428716e-12 P( X = 12 ) = 9.524063e-14 P( X = 13 ) = 1.156769e-15 P( X = 14 ) = 8.69751e-18 P( X = 15 ) = 3.051758e-20 P( X ≥ 2 ) = 1 - P(X = 0) - P(X =1) = 0.1709525