Please help me showing minimum work ASAP, will rate A+++++ Question 1. (a) A lot
ID: 2956781 • Letter: P
Question
Please help me showing minimum work ASAP, will rate A+++++Question 1.
(a) A lot of 50 refrigerators contains 30 refrigerators which are defective.
Suppose 3 refrigerators are randomly selected from this lot, without replacement.
i. Find the probability that all three selected refrigerators are defective.
ii. Find the probability that the third selected refrigerator is defective if the first
two selected refrigerators are defective.
iii. With no information as to whether the first two selected refrigerators are
defective or not, find the probability that the third selected refrigerator is
defective.
(b) Suppose we again select refrigerators from the lot in (a) at random, without
replacement. In order to have the probability that all the selected refrigerators
are non-defective to be less than 0.1, what is the minimum number of refrigerators
that we need to select?
(c) Suppose we select refrigerators from another lot. This lot is large enough that
we can assume that our sampling is done with replacement. In this lot, 60%
of refrigerators are defective. In order to have the probability that at least one
selected refrigerator is non-defective to be at least 0.9, what is the minimum
number of refrigerators that we need to select?
Explanation / Answer
a.i. this would be 30/50 for the first refrigerator pulled, 29/49 for the second, and 28/48 for the third. we just multiply all three of these fractions together to get .20714, which is 20.71% probability that all 3 are defective. a.ii. this is 28/48 which is 7/12 or .58333 or 58.33% a.iii. There are a number of ways to interpret this depending on the method or the instructor. You can either ignore the pulled 2 and simply use the 30 possible defectives to be in the 48 remaining, or you can use the 30/50 ratio to decrease the possibilities for each one that has been pulled. The first choice would simply be 30/48 which is 5/8 or .625 b. You would need to use the 20/50 ratio to work with, as there are 20 good since there are 30 bad in the lot of 50. pulling two would be 20/50 * 19/49 = .1551 pulling three would be 20/50 * 19/49 * 18/48 = .0581 three would be under .1, at .0581 c. this uses a .6 bad probability and .4 good probability A 0.9 probability would not be possible as this is over the 0.4 probability inherent to the problem Perhaps it is asking for .09? If so, you would need to pull three, because .4 * .4 * .4 = .064