Chances of A Certain Number of D's as Multiple Choice Answers

Question

In a test I took recently, 36 of the 90 answers were D. The options were A, B, C, and D. What are the chances that this happened randomly, and was not the result of a bias by the test creator?

Answer 1

You will want to look up information about the binomial coefficient ${n\choose k}$ (read '$n$ choose $k$'). It is the number of ways to choose $k$ objects from a collection of $n$ objects. It has the formula $$ {n\choose k}=\frac{n!}{(n-k)!k!}.$$

In this case $n$ would be $90$, the number of questions. The number of questions we want to be the same character is $36$ so that would be $k$.

Now we can calculate the answer. The $54$ questions that don't make up the $33$ can be one of three types so we need to multiply our answer by $3^{54}$. Furthermore the character type of the $36$ questions can be one of four types so we need to multiply our answer by $4$.

This gives us $$ \frac{4\cdot 3^{54}\cdot {90\choose 36}}{4^{90}} \approx 2.63\times 10^{-3}. $$

Answer 2

This concerns binomial distribution. There are four conditions that must be satisfied for a binomial distribution:

$1.$Binary? The possible outcomes of each trial can be classified as "success" or "failure".

$2.$Independent? Trials must be independent; knowing the result of one trial must not have any effect on the result of any other trial.

$3.$Number? The number of trials n of the chance process must be fixed in advance.

$4.$Success? On each trial, the probability p of success must be the same.

Lets check these four conditions now.

$1.$Binary? In your question, success will be getting a "D", and failure will be not getting a "D". $\color{green}\checkmark$

$2.$ Independent? The trials are independent (whether the previous answer is D or not has no effect on the answer to the next question). $\color{green}\checkmark$

$3.$ Number? The number of trials is fixed at 90. $\color{green}\checkmark$

$4.$ Success? The chance of getting a "D" is always $0.25$. $\color{green}\checkmark$

Since all four conditions are satisfied, binomial distribution applies here.

Now we can just apply the formula for the distribution.

$$P(X=36)=\binom{90}{36}(0.25)^{36}(1-0.25)^{90-36}\approx 6.56\times10^{-4}$$