Question

A multiple choice test question has five altemative answers, of which only one is correct. If a student has done his home work, then he is sure to identify the correct answer; otherwise, he chooses an answer at random.
Let $E$ : denotes the event that a student does his home work with $P(E)=p$
and $F$ : denotes the event that he answer the question correctly.
Suppose that each question has $n$ alternative answers of which only one is correct, and $p$ is fixed but not equal to 0 or 1 then $P(E/F)$

• A. decreases as $n$ increases for all $p\in (0,1)$
• B. increases as $n$ increases for all $p\in (0,1)$
• C. remains constant for all $p\in (0,1)$
• D. decreases if $p\in (0,0.5)$ and increases if $p\in (0.5,1)$ as $n$ increases

Answer 1

Answer: (B)

$P(E)=p$
$P(F)=P(E\cap F)+P(\bar{E}\cap F)$
$P(F)=P(E)P(F/E)+P(\bar{E})P(F/E)$
$=p\cdot 1+(1-p)\cdot \frac{1}{5}=\frac{4p}{5}+\frac{1}{5}=\frac{4p+1}{5}$

If each questions has n alternatives then
$P(F)=p+(1-p)\frac{1}{n}=p\left(1-\frac{1}{n}\right)+\frac{1}{n}=\frac{(n-1)p+1}{n}$
$\therefore P(E/F)=\frac{np}{(n-1)p+1}$ which increases as $n$ increases for a fixed $p\Rightarrow$