Many manufacturers have quality control programs that include inspection of incoming materials for defects. Suppose a computer manufacturer receives computer boards in lots of five. Two boards are selected from each lot for inspection. We can represent possible outcomes of the selection process by pairs. For example, the pair (1, 2) represents the selection of boards 1 and 2 for inspection.
a. List the ten different possible outcomes.
b. Suppose that boards 1 and 2 are the only defective boards in a lot of five. Two boards are to be chosen at random. Define X to be the number of defective boards observed among those inspected. Find the probability distribution of X.
c. Let F(x) denote the cdf of X. First determine(0)=P( X ≤ 0), F(1), and F(2); then obtain F(x) for all other x.

Answer:

Hence, the list ten different possible outcomes are [tex](1,2),(1,3),(1,4),(1,5),(2,3),(2,4),(2,5),(3,4),(3,5),(4,5)[/tex], the probability distribution of X are [tex]0.3,0.6,0.1[/tex] and the cumulative distribution function is [tex]F(x)=\left\{\begin{array}{lc}0 & x<0 \\0.3 & 0 \leq x<1 \\0.9 & 1 \leq x<2 \\1 & 2 \leq x\end{array}\right.[/tex]

Step-by-step explanation:

(a)

List the ten different possible outcomes.

The number of ways that two boards are drawn from each lot of [tex]5[/tex] lots are [tex]{5 \choose 2}[/tex]

    [tex]{5 \choose 2}=\frac{5!}{2!(5-2)!}[/tex]

[tex]\Rightarrow {5 \choose 2}=\frac{5\times 4\times 3!}{2!\times 3!}[/tex]

[tex]\Rightarrow {5 \choose 2}=\frac{5\times 4}{2!}[/tex]

[tex]\Rightarrow {5 \choose 2}=\frac{20}{2}[/tex]

[tex]\Rightarrow {5 \choose 2}=10[/tex]

So, the [tex]10[/tex] combinations are as follows :

[tex](1,2),(1,3),(1,4),(1,5),(2,3),(2,4),(2,5),(3,4),(3,5),(4,5)[/tex]

(b)

The aim is to find the probability distribution when two boards are chosen at random and board's one and two is the only defective boards.

Let X be the number of defective boards in the lot.

Therefore, select the combinations of the boards that are without 1 and 2 from 10 combinations.

Compute the [tex]P(X=0)[/tex]

 [tex]P(X=0)=P\{(3,4),(3,5),(4,5)\}[/tex]

                 [tex]=\frac{3}{10}[/tex]

                 [tex]=0.3[/tex]

The probability that defectives did not occur in the lot is 0.3

Compute the [tex]P(X=1)[/tex]

Select the combinations of the boards that are with one defective either 1 and 2 from the 10 combinations.

[tex]P(X=1)=P\{(1,3),(1,4),(1,5),(2,3),(2,4),(2,5)\}[/tex]

                [tex]=\frac{6}{10}[/tex]

                [tex]=0.6[/tex]

The probability that one defectives occurred in the lot is 0.6.

Compute the [tex]P(X=2)[/tex]

Select the combinations of the boards that are with two defective 1 and 2 from the 10 combinations.

[tex]P(X=2)=P\{(1,2)\}[/tex]

                [tex]=\frac{1}{10}[/tex]

                [tex]=0.1[/tex]

The probability that two defectives occurred in the lot is 0.1.

(c)

The cumulative distribution function (cdf)[tex]F(X=x)[/tex]  is defined as,

[tex]\begin{aligned}F(X&=x)=P(X \leq x) \\F(0) &=P(X \leq 0) \\&=P(X=0) \\&=0.3 \\F(1) &=P(X \leq 1) \\&=P(X=0)+P(X=1) \\&=0.3+0.6 \\&=0.9 \\F(2) &=P(X \leq 2) \\&=P(X=0)+P(X=1)+P(X=2) \\&=0.3+0.6+0.1 \\&=1.0\end{aligned}[/tex]

Therefore, the cumulative distribution function (cdf) [tex]F(X=x)[/tex] is,

[tex]F(x)=\left\{\begin{array}{lc}0 & x<0 \\0.3 & 0 \leq x<1 \\0.9 & 1 \leq x<2 \\1 & 2 \leq x\end{array}\right.[/tex]

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