Probability of number of random answers being correct.

Question

My teacher gave us a true-false test with 100 "questions." Only there were no questions. He had an answer key and was trying to prove that if we answered true and false questions randomly, the class average would be around 50%. I got 7 right out of 100, so, of course, 93 wrong out of 100. He said that he had never had a score that far from the mean.

What is the probability of getting only 7 true-false questions out of 100 correct?

So far, life has proven you right. If only those numbers had been reversed. — Laura, Dec 26 '13 at 05:14

score 4 · Answer 1 · answered Dec 26 '13 at 03:15

4

Very small. To get at most 7 right, $$ \sum_{k= 0}^7 {100\choose k}{1\over 2^{100}}.$$ This is a wee-tiny number.

answered Dec 26 '13 at 03:15

ncmathsadist

49,383

Approximately $1.363 \times 10^{-20}$. – xzackli Dec 26 '13 at 03:17
Remember ${n\choose k}$ is defined to be $\frac{n!}{k!(n-k)!}$. – Sak Dec 26 '13 at 03:17
Zack, that's wee-tiny. – ncmathsadist Dec 26 '13 at 03:19
Thank you everyone. With a little help, I understood all this. – Laura Dec 26 '13 at 05:12

score 1 · Accepted Answer · answered Dec 26 '13 at 09:05

The number of correct guesses is governed by something called the binomial distribution.

$$P(k\,|\,n,p) = \left(\begin{array}{c}n\\k\end{array}\right)p^k(1-p)^{n-k}$$

The binomial distribution gives the probability of observing one of two possible outcomes (for example, a coin toss coming out heads or tails, a student guessing a true/false question right) $k$ times out of $n$ when there is a probability of $p$ of getting the desired outcome on any one trial.

In your case $p = \frac{1}{2}$ since students are randomly guessing the answer to true/false questions and are as likely to get them right as they are to get them wrong. You had a total of $n=100$ questions to answer and got $k=7$ of them right. The probability of getting exactly $7$ right is

$$ \begin{align} P(k=7\;|\;n=100,\,p=1/2) &= \left(\begin{array}{c}100\\7\end{array}\right)2^{-100} \\ &\approx 1.26 \times 10^{-20} \end{align} $$

The probability of guessing at most $7$ right is

$$ \begin{align} P(k\le7\;|\;n=100,\,p=1/2) &= \sum_{i=0}^7\left(\begin{array}{c}100\\i\end{array}\right)2^{-100} \\ &\approx 1.36 \times 10^{-20} \end{align} $$

which is only a little greater than the probability of getting exactly $7$ right.

This tells you, as an individual student, that it is extremely unlikely to get such a low score. For your teacher, who has asked many students to do this experiment, the probability of having seen such an outcome is still very small. It's approximately $s \times 1.36 \times 10^{-20}$. Even for a thousand students this is still of the order of $10^{-17}$, which is still a ridiculously small number.

Probability of number of random answers being correct.

2 Answers2