Chapter 17, Problem 16E

(a)

To determine

To find: The researcher should predict to find a colour-blind man by the number of persons.

Answer to Problem 16E

12

Explanation of Solution

Given:

Male colour blind is 0.08 probability. It is geometric model

Suppose X indicates how many colour blindness failures have happened since the first male.

  $p=0.08$

Formula used:

  $E\left(X\right)=\frac{\left(1-p\right)}{p}$

Calculation:

  $\begin{array}{c}P\left(X=x\right)={\left(1-p\right)}^{x}p,\\ x=0,1,\mathrm{2...}\\ p=0.08\end{array}$

The number of males predicted to be tested before identifying a blind first color is,

  $\begin{array}{c}E\left(X\right)=\frac{\left(1-p\right)}{p}\\ =\frac{1-0.08}{0.08}\\ =\frac{0.92}{0.08}\\ =11.5\end{array}$

Until first finding blind man, about 12 males should be tested.

(b)

To determine

To find: The probability that no one of the first 4 men researchers would find any one colour-blind.

Answer to Problem 16E

0.7164

Explanation of Solution

Given:

Male colour blind is 0.08 probability. It is geometric model

Suppose X indicates how many colour blindness failures have happened since the first male.

  $p=0.08$

Calculation:

The probability of not being blind to a single color among the first four men. This means that before the first success there are four trials. Therefore,

  $\begin{array}{c}P\left(X=4\right)={\left(0.92\right)}^4\\ =0.716393\\ =0.7164\end{array}$

There is also a probability that 71.64% of the first four men would find no single-color blinds.

(c)

To determine

To find: The probability will be tested by the 6th person, the first colour-blind man checked.

Answer to Problem 16E

0.05273

Explanation of Solution

Given:

Male colour blind is 0.08 probability. It is geometric model

Suppose X indicates how many colour blindness failures have happened since the first male.

  $p=0.08$

Calculation:

It is possible the sixth person would be the first blind man to be identified. This means that before the first success there are five trials. Therefore,

  $\begin{array}{c}P\left(X=5\right)={\left(0.92\right)}^5\left(0.08\right)\\ =0.659082\times 0.08\\ =0.052727\\ =0.05273\end{array}$

Thus, the first blind man seen in colours, 5.27%, is the sixth man.

(d)

To determine

To find: The possibility that the researcher will identify someone who is colour-blind before the 10th man is tested.

Answer to Problem 16E

0.5656

Explanation of Solution

Given:

Male colour blind is 0.08 probability. It is geometric model

Suppose X indicates how many colour blindness failures have happened since the first male.

  $p=0.08$

Calculation:

She would possibly find someone who is blind in colour before the 10th person checks,

  $\begin{array}{c}P\left(X\le 9\right)={\displaystyle \sum _{x=0}^{9}P\left(X=x\right)}\\ =P\left(X=0\right)+P\left(X=1\right)+P\left(X=2\right)+\mathrm{....}+P\left(X=9\right)\\ ={\left(0.92\right)}^0\left(0.08\right)+{\left(0.92\right)}^1\left(0.08\right)+{\left(0.92\right)}^2\left(0.08\right)+\mathrm{....}+{\left(0.92\right)}^9\left(0.08\right)\\ =0.08\left[{\left(0.92\right)}^0+{\left(0.92\right)}^1+{\left(0.92\right)}^2+\mathrm{....}+{\left(0.92\right)}^9\right]\\ =0.08\left[\frac{1-{\left(0.92\right)}^{9+1}}{1-0.92}\right]\left(\text{since, 1+}x+x^2+\mathrm{...}+x^n=\frac{1-x^{n+1}}{1-x}\right)\\ =0.08\left[1+{\left(0.92\right)}^1+{\left(0.92\right)}^2+\mathrm{....}+{\left(0.92\right)}^9\right]\\ =0.08\left[\frac{1-{\left(0.92\right)}^{9+1}}{1-0.92}\right]\\ =0.08\left[7.070144\right]\\ =0.565612\end{array}$

She may find someone who is blind until the 10th person is tested at 56.56 percent.