Question

From a lot of 15 , a sample of 3 is selected. How many different sample combinations are possible? From a lot of 60 a sample of 5 is selected. How many different permutations are possible? If the probability of obtaining 1 non conforming unit in a sample of 2 is .20 and the probability of 2 non conforming units is .25 , what is the probability of 0 non conforming units?

Answer 1

The number of sample combinations when selecting 3 from a lot of 15 is 455. The number of permutations for selecting 5 from a lot of 60 is 54615120. The probability of selecting 0 non-conforming units, given the other probabilities, is 0.55.

Step-by-step explanation:

To find how many different sample combinations are possible when selecting 3 from a lot of 15, we use the combination formula which is C(n, k) = n! / (k!(n-k)!), where n is the total number of items, k is the number of items to choose, and ! denotes factorial. Hence, the number of combinations is C(15, 3) = 15! / (3!12!) = 455.

For the second part, we consider the number of different permutations. The permutation formula is P(n, k) = n! / (n-k)!, therefore the number of permutations for selecting 5 from a lot of 60 is P(60, 5) = 60! / (60-5)! = 60! / 55! which equals 54615120.

Concerning probability, we have two probabilities given for 1 and 2 non-conforming units. To find the probability of 0 non-conforming units, we use the complement rule: P(0) = 1 - (P(1) + P(2)) = 1 - (0.20 + 0.25) = 0.55.