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{4 p}{5}+\frac{1}{5}=\frac{4 p+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 $p \Rightarrow$