Chapter 5.3, Problem 99E

a.

To determine

To find:the percentage of calls results in a contribution.

Explanation of Solution

Given:

The given table shows that the presence of antibodies to HIV.

	Test positive	Test negative
Antibodies present	0.9895	0.0015
Antibodies absent	0.0006	0.994

Calculation:

The table contains the probability of getting positive/negative test result when we know that there are antibodies present or antibodies absent.

Also, it is known that 1% of the population carries antibodies.

  

b.

To determine

To find: the percentage who contribute are recent donor.

Answer to Problem 99E

  $P\left(P\right)=1.5925\%$

Explanation of Solution

Given:

The given table shows that the presence of antibodies to HIV.

	Test positive	Test negative
Antibodies present	0.9895	0.0015
Antibodies absent	0.0006	0.994

Calculation:

Let

  $\begin{array}{l}A=\text{Anti-bodies pesent,}{A}^C=\text{Anti-bodies absent}\\ P=\text{Positive,}{P}^C=\text{negative}\end{array}$

Use the complemenr rule:

  $P\left(A^C\right)=1-P\left(A\right)=1-0.01=0.99$ .

Use the general multiplication rule:

  $\begin{array}{l}P\left(P\text{ and }A\right)=P\left(A\right)\times P\left(P|A\right)=0.01\times 0.9985=0.009985\\ P\left(P\text{ and }{A}^C\right)=P\left(A^C\right)\times P\left(P|A^C\right)=0.99\times 0.006=0.00594\end{array}$

Use the addition rule mutually exclusive events (as the vechiles cannot be filled with two types at the same time).

  $\begin{array}{l}P\left(P\right)=P\left(P\text{ and }A\right)+P\left(P\text{ and }{A}^C\right)\\ =0.009985+0.00549\\ =0.015925=1.5925\%\end{array}$

c.

To determine

To find: the percentage who contribute are recent donor.

Answer to Problem 99E

  $P\left(A|P\right)=0.6270$

Explanation of Solution

Given:

The given table shows that the presence of antibodies to HIV.

	Test positive	Test negative
Antibodies present	0.9895	0.0015
Antibodies absent	0.0006	0.994

Concept used:

Definition conditional probability:

  $P\left(A|B\right)=\frac{P\left(A\text{ and }B\right)}{P\left(B\right)}$

Calculation:

Let

  $\begin{array}{l}A=\text{Anti-bodies pesent,}{A}^C=\text{Anti-bodies absent}\\ P=\text{Positive,}{P}^C=\text{negative}\end{array}$

From pair (b), $P\left(P\right)=1.5925\%$

Definition conditional probability:

  $P\left(A|B\right)=\frac{P\left(A\text{ and }B\right)}{P\left(B\right)}$

Thus obtained:

  $\begin{array}{l}P\left(A|P\right)=\frac{P\left(A\text{ and }P\right)}{P\left(P\right)}\\ =\frac{0.009985}{0.015925}=0.6270\end{array}$