Question

Suppose that a batch of 100 items contains 6 that are defective and 94 that are not defective. If X is the number of defective items in a randomly drawn sample of 10 items from the batch, find (a) P{X=0} and (b) P(X>2}.

Answer 1

Answer:

a) P(X = 0) = 0.5223

b) P(X > 2) = 0.0125

Step-by-step explanation:

A probability is the number of desired outcomes divided by the number of total outcomes.

The order that the items are selected is not important, so the combinations formula is used to solve this problem.

Combinations formula:

$C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

In this problem, we have that

100 items

6 defective

94 not defective.

a) P{X=0}

None defective.

Desired outcomes

Combinations of 10 from a set of 94. So

$D = C_{94,10} = \frac{94!}{10!84!} = 9041256800000$

Total outcomes:

Combinations of 10 from a set of 100. So

$T = C_{100,10} = \frac{100!}{10!90!} = 17310309000000$

P(X = 0)

$p = \frac{D}{T} = \frac{9041256800000}{17310309000000} = 0.5223$

(b) P(X>2}.

Either two or less are defective, or more than two are defective. The sum of the probabilities of these events is decimal 1. So

$P(X \leq 2) + P(X >2) = 1$

$P(X > 2) = 1 - P(X \leq 2)$

In which

$P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2)$

P(X = 0)

$P(X = 0) = \frac{D}{T} = \frac{9041256800000}{17310309000000} = 0.5223$

P(X = 1)

Desired outcomes

Combinations of 9 from a set of 94(non defecive) and 1 from a set of 6(defective). So

$D = C_{84,9}*C_{6,1} = \frac{94!}{9!85!}*\frac{6!}{1!5!} = 6382063700000$

Total outcomes:

Combinations of 10 from a set of 100. So

$T = C_{100,10} = \frac{100!}{10!90!} = 17310309000000$

P(X = 1)

$P(X = 1) = \frac{D}{T} = \frac{6382063700000}{17310309000000} = 0.3687$

P(X = 2)

Desired outcomes

Combinations of 8 from a set of 94(non defecive) and 2 from a set of 6(defective). So

$D = C_{94,8}*C_{6,2} = \frac{94!}{8!86!}*\frac{6!}{2!4!} = 1669726000000$

Total outcomes:

Combinations of 10 from a set of 100. So

$T = C_{100,10} = \frac{100!}{10!90!} = 17310309000000$

P(X = 2)

$P(X = 2) = \frac{D}{T} = \frac{1669726000000}{17310309000000} = 0.0965$

Finally

$P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2) = 0.5223 + 0.3687 + 0.0965 = 0.9875$

$P(X > 2) = 1 - P(X \leq 2) = 1 - 0.9875 = 0.0125$