Probability in multiple choice exams

• A multiple choice exam has 175 questions.
• Each question has 4 possible answers.
• Only 1 answer out of the 4 possible answers is correct.
• The pass rate for the exam is 70% (123 questions must be answered correctly).

• We know 100 questions were answered correctly.

• What is the probability of passing the exam, if one were to guess on the remaining 75 questions?

The number $X$ of correct answers in the remaining $75$ that were answered randomly, is binomially distributed with parameters $n=75$ and $p=0.25$, in symbols $$X \sim B(n=75, p=0.25)$$ The probability to pass the exam is equal to $$P(X\ge23)=\sum_{x=23}^{75}P(X=x)$$ Since it is difficult to calculate the above sum if you are not allowed to use a calculator or a program like excel, you can approximate $X$ with a normal random variable $Y$ such that $$Y \sim N(μ=np, σ^2=np(1-p)) \implies Y \sim N(μ=18.75, σ^2=14.065)$$ Now $$P(X\ge 23)=1-P(Y\le 22.5)$$ where we applied also the continuity correction.

• Thanks Stefanos. Can you solve the equation as well to derive the answer? – Brittany Nov 4 '14 at 14:15
• You are welcome Brittany. No, I would suggest that you try to take it by yourself from here. Good luck.... – Jimmy R. Nov 4 '14 at 14:26

We have already answered 100 questions, so there are only 75 questions left to answer. Since we are guessing our way through the multiple choice questions, our probability of success in each question will be $\frac{1}{4}$ Since the pass mark is $\frac{123}{75}$, we need at least $\frac{23}{75}$ in the final 75 questions. This is the same as saying that we need to find $P(X\ge23)$, i.e. "What is the probability of getting 23 or more questions correct?"

The information we have so far suggests that we can use the binomial distribution. $X \sim B(n,p)$. Where $n=75$ and $p=\frac{1}{4}$ in your question. However, we may have a slight problem. 75 is too large for us to use the $ncr$ formula and binomial tables don't generally include $n=75$. Unless you have a graphical calculator or some sort of statistical software, we will need to use a $normal \ approximation$ in order to answer your question.

When do you need to normally approximate?

• Look at $np$ and $nq$. $\left(n=75 \ and \ p=\frac{1}{4} \right)$
• Look at n, is it "Large"? ($n\ge30$ is normally a candidate).
• if $n$p and $nq>5$ and $n$ is large, you can try a normal approximation.
• Also, if $p$ is close to $\frac{1}{2}$, this is another indication.

$X \sim B \left(75,\frac{1}{4} \right)$

we are looking for $P(X\ge23)$

Normally approximating:

$Y \sim N(18.75,14.0625)$ as $Y \approx N(np,npq)$ is the normal approximation to $X \sim B(n,p)$ where $q=1-p$

Applying continuity correction, $P(X \le 22)$ becomes $P(Y \le 22.5)$

Normal distribution: $Z = \frac{X-\mu}{\sigma}$

$Z= \frac{22.5-18.75}{\sqrt14.0625}$

We need to find $1-P(Z<1)$

Look for $\varPhi(1)$ on normal tables, the answer is $0.84134$.

Therefore, the normal approximation gives us an answer of $1-0.84134=0.15866$

The normally approximated answer to your question is $0.159(3dp)$.

Continuity correction:

$P(X\ge23)$ is the same as $1-P(X\le22)$

$P(X\le a)$ becomes $P(Y\le a+0.5)$ after the continuity correction.