Hi Binomial Probability: n = 30, p = 3/30 = .1 what then is the probability that the inspector's sample contains: Using TI or similarly an inexpensive calculator like an Casio fx-115 ES plus P( x = 0) = .0424 0rP( x = 0) = 1(.1^0)(.9^30) = .0424 ----------------- n = 3, p = .8 P(x = 0) = .008 P(at least one) = 1 - P(0) == 1 - .008 = .992 Wish You the Best in your Studies.
(i) The probability in this problem is the ratio of two numbers. The denominator is the number of all possible triples of batteries from the whole set of 30 batteries. The number of such triples is the number of all combinations= = 4060. The numerator of this fraction is the number of all triples comprising of good batteries. It is the number of combinations = = = 1140. Thus the probability P(i) = = = = 0.2808 (rounded). ANSWER (ii) In this problem, the probability is the COMPLEMENT to what we found in (i) P(ii) = 1 - P(i) = 1 - 0.2808 = 0.7192 (rounded). ANSWER