In a lot of 10 components, 2 are sampled at random for inspection. Assume that in fact exactly 2 of the 10 components in the lot are defective. Let X be the number of sampled components that are defective.

a. Find P(X = 0).

b. Find P(X = 1).

c. Find P(X = 2).

d. Find the probability mass function of X.

e. Find the mean of X.

f. Find the standard deviation of X.

Solutions:

Step of 1 of 5:

In a lot of 10 components, 2 are sampled at random for inspection. Here exactly 2 of the 10 components in the lot are defective. Let X be the number of defectives. Then we have to find

P(X=0)P(X=1)P(X=2)Probability mass function of X.Mean of XStandard deviation of XStep 2 of 5:

(a)

Here it is given that in a lot 10 components exactly two components are defectives. That means the remaining 8 are non-defectives. 2 are sampled at random for inspection. Since X is the number of defectives

P(X=0)= P(randomly selected two components are non - defectives)

The total ways to select two non- defective components are = 8C2

And the total ways to select two components out of 10 components are = 10C2

Therefore

P(X=0) =

=

= 0.6222

Therefore the probability P(X=0) 0.6222 .

(b) We have to find the probability that P(X=1)

P(X=1) = P(that one of the two selected components is defective and the other one is non-defective.)

The total number of ways to select one defective item and one non-defective item is

= 8C1 2C1 .

The total ways to select two components out of 10 components are = 10C2 .

Therefore ,

P(X=1) =

= 16/45

= 0.3555.

Therefore the probability P(X=1) = 0.3555.

Step 3 of 5:

(c) We have to find the probability that P(X=2)

P(X=2) = P(that two selected components are defective.)

The total number of ways to select two defective components are = 2C2.

The total ways to select two components out of 10 components are = 10C2 .

Therefore ,

P(X=1) =

= 1/45

= 0.022.

Therefore the probability P(X=2) = 0.022.