Chapter PIV, Problem 21RE

(a)

To determine

To find: the probability that researcher would pass the statistics.

Answer to Problem 21RE

0.7167

Explanation of Solution

Given:

  $\begin{array}{c}P\left(A/S\right)=0.8\\ P\left(A/K\right)=0.6\end{array}$

Formula used:

  $P\left(A\right)=P\left(A/S\right)P\left(S\right)+P\left(A/K\right)P\left(K\right)$

Calculation:

Suppose

A = student passes statistics

S = student gets professor Scedastic

K = student gets professor kurtosis

It is given that

  $\begin{array}{c}P\left(S\right)=\frac{70}{120}\\ P\left(K\right)=\frac{50}{120}\end{array}$

The probability of transferring the statistics to the researcher is

  $\begin{array}{c}P\left(A\right)=P\left(A/S\right)P\left(S\right)+P\left(A/K\right)P\left(K\right)\\ =\left(0.8\right)\frac{70}{120}+\left(0.6\right)\frac{50}{120}\\ =0.7167\end{array}$

There is 71.67 possibility that researcher would pass.

(b)

To determine

To find: the probability that researcher will get the professor.

Explanation of Solution

Formula used:

  $P\left(K|\overline{A}\right)=\frac{P\left(K\cap \overline{A}\right)}{P\left(\overline{A}\right)}$

Calculation:

The required probability is

  $P\left(K|\overline{A}\right)=\frac{P\left(K\cap \overline{A}\right)}{P\left(\overline{A}\right)}$

  $\begin{array}{c}P\left(K\cap \overline{A}\right)=P\left(K\right)P\left(\overline{A}|K\right)\\ =P\left(K\right)\left[1-P\left(A|K\right)\right]\\ =\frac{50}{120}\left[1-0.6\right]\\ =0.1667\end{array}$

  $\begin{array}{c}P\left(\overline{A}\right)=1-P\left(A\right)\\ =1-0.7167\\ =0.2833\end{array}$

Substituting the values get

  $\begin{array}{c}P\left(K|\overline{A}\right)=\frac{P\left(K\cap \overline{A}\right)}{P\left(\overline{A}\right)}\\ =\frac{0.1667}{0.2833}\\ =0.5882\end{array}$

Thus, the required probability is 0.5882.