Question 697630
The distribution of the number of defective items in this run is binomial with n = 75 items and p = .05 of each item being defective. The probability mass function of the binomial distribution is:<br>

{{{P(X = k) = (nCk)*(p^k)*(1-p)^(n-k)}}}, where nCk is the number of combinations of k objects chosen from n = {{{n!/(k!*(n-k)!)}}}<br>

1) In this case, we set k = 5. n = 75 and p = .05, so:<br>

{{{P(X = 5) = (75C5)*(.05^5)*(.95)^(70)}}}
{{{P(X = 5) = (75!/(5!*(70)!))*(.05^5)*(.95)^(70)}}}
{{{P(X = 5) = (75!/(5!*(70)!))*(.05^5)*(.95)^(70) = .1488}}}<br>

2) In this case, we set k = 0. The binomial probability mass function reduces to {{{(1-p)^n}}} when k = 0, so P(X = 0) = {{{(.95)^75 = .0213}}}<br>

3) The event that at least one item is defective is the compliment of the event that there are no defective items. The probability of a complimentary event happening is 1 - P(original event), so the probability of at least one defective item is 1 - .0213 = .9787.