Q. 4.45

Question

A student is getting ready to take an important oral examination and is concerned about the possibility of having an “on” day or an “off” day. He figures that if he has an on the day, then each of his examiners will pass him, independently of one another, with probability $8$ , whereas if he has an off day, this probability will be reduced to $4$ . Suppose that the student will pass the examination if a majority of the examiners pass him. If the student believes that he is twice as likely to have an off day as he is to have an on the day, should he request an examination with $3$ examiners or with $5$ examiners?

Step-by-Step Solution

Verified

Answer

The probability of passing is more, if $3$ examiners, So he should prefer $3$ examiners.

1Step1: Given Information

A student is getting ready to take an important oral examination and is concerned about the possibility of having an “on” day or an “off” day.

Probability of off day $= \frac{2}{3}$

Probability of on day $= \frac{1}{3} .$

2Step 2: Solution of the Problem

If he request $3$ Examiners then:

$P [pass if 2 or more emaminers pass him]$

$= \frac{2}{3} [(\begin{array}{l} 3 \\ 2 \end{array}) (0.4)^{2} \cdot (0.6) + (\begin{array}{l} 3 \\ 3 \end{array}) (0.4)^{3}] + \frac{1}{3} [(\begin{array}{l} 3 \\ 2 \end{array}) (0.8)^{2} \cdot (0.2) + (\begin{array}{l} 3 \\ 3 \end{array}) (0.8)^{3}]$

$= \frac{2}{3} [0.288 + 0.064] + \frac{1}{3} [0.384 + 0.512]$

We get,

$= 0.2347 + 0.2987$

$= 0.533 .$

3Step 3: Computation of the Probability

If he request $5$ Examiners then:

$P [pass if 3 or more examiners pass him]$

$P (pass) = \frac{2}{3} [(\begin{array}{l} 5 \\ 3 \end{array}) (0.4)^{3} \cdot (0.6)^{2} + (\begin{array}{l} 5 \\ 4 \end{array}) (0.4)^{4} \cdot (0.6) + (\begin{array}{l} 5 \\ 5 \end{array}) (0.4)^{5}]$

$+ \frac{1}{3} [(\begin{array}{l} 5 \\ 3 \end{array}) (0.8)^{3} \cdot (0.2)^{2} + (\begin{array}{l} 5 \\ 4 \end{array}) (0.8)^{4} \cdot (0.2) + (\begin{array}{l} 5 \\ 5 \end{array}) (0.8)^{5}]$