Chapter 17, Problem 15E

(a)

To determine

To find: To find someone of Type AB blood, the numbers of donors must be confirmed.

Answer to Problem 15E

25

Explanation of Solution

Given:

p the probability of a individual getting blood type AB $p=0.04$ .

X is the AB blood donor already identified $x^{th}$ trail.

  $X~\text{Geometric}\left(0.04\right)$

Formula used:

  $E\left(X\right)=\frac{1}{p}$

Calculation:

The expected value is,

  $\begin{array}{c}E\left(X\right)=\frac{1}{p}\\ =\frac{1}{0.04}\\ =25\end{array}$

The estimated number of persons to check for type AB blood is also 25.

(b)

To determine

To find: The possibility of the Type AB donor checking among the 1^st 5 individuals.

Answer to Problem 15E

0.185

Explanation of Solution

Given:

p the probability of a individual getting blood type AB $p=0.04$ .

X is the AB blood donor already identified $x^{th}$ trail.

  $X~\text{Geometric}\left(0.04\right)$

Calculation:

It is possible that among the first five persons examined there is an AB type donor,

  $\begin{array}{c}P\left(X=1\right)+P\left(X=2\right)+P\left(X=3\right)+P\left(X=4\right)+P\left(X=5\right)\\ =\left(0.04\right)+\left(0.96\right)\left(0.04\right)+{\left(0.96\right)}^2\left(0.04\right)+{\left(0.96\right)}^3\left(0.04\right)+{\left(0.96\right)}^4\left(0.04\right)\\ =0.04+0.038+0.037+0.035+0.034\\ =0.185\end{array}$

It is also likely that 18.5% of the first five persons tested would have a type AB donor.

(c)

To determine

To find: The probability of the first Type AB donor in the first six individuals being identified.

Answer to Problem 15E

0.2172

Explanation of Solution

Given:

p the probability of a individual getting blood type AB $p=0.04$ .

X is the AB blood donor already identified $x^{th}$ trail.

  $X~\text{Geometric}\left(0.04\right)$

Calculation:

In the first six people tested, the likelihood is that the donor type AB is identified

  $\begin{array}{c}P\left(X=1\right)+\mathrm{...}+P\left(X=6\right)=\left[\begin{array}{l}\left(0.04\right)+\left(0.96\right)\left(0.04\right)+{\left(0.96\right)}^2\left(0.04\right)+\\ {\left(0.96\right)}^3\left(0.04\right)+{\left(0.96\right)}^4\left(0.04\right)+{\left(0.96\right)}^5\left(0.04\right)\end{array}\right]\\ =\left[0.04+0.0384+0.0368+0.0352+0.034+0.0328\right]\\ =0.2172\end{array}$

There is also a 21.72% probability that the AB donor is among the first persons to be tested.

(d)

To determine

To find: The probability of not having a donor of Type AB before the tenth individual.

Answer to Problem 15E

0.693

Explanation of Solution

Given:

p the probability of a individual getting blood type AB $p=0.04$ .

X is the AB blood donor already identified $x^{th}$ trail.

  $X~\text{Geometric}\left(0.04\right)$

Calculation:

Let Y be an AB blood donor not previously identified in the trail.

The probability that the blood donor type AB was not identified until the 10th person is,

  $\begin{array}{c}P\left(Y=10\right)={\left(0.96\right)}^{10-1}\\ =0.693\end{array}$

The likelihood that the blood donor type AB was unknown until the 10th individual is therefore 0.693