# Probability that you have exactly one correct answer on a test

I have two questions from a practice test which I have some concerns about:

Assume you write a multiple-choice exam that consists of 100 questions. For each question, 4 options are given, one of which is the correct one. If you answer each of the 100 questions by choosing an answer uniformly at random, what is the probability that you have exactly one correct answer?

(a) $\frac{100}{4^{100}}$
(b) $\frac{3^{99}}{4^{100}}$
(c) $\frac{100+3^{99}}{4^{100}}$
(d) $\frac{100\cdot3^{99}}{4^{100}}$

The answer is (d). Can someone help me understand how to go about this problem?

• How does $100\cdot3^{99}$ count the number of ways there can be exactly one correct answer?
• I know that $3^{99}$ is the number of ways to choose $3$ possible answers from $99$ questions, while $1$ question is already fixed (correct). Why multiply by $100$?

You flip a fair coin 5 times. Define the events

$A =$ "the number of heads is odd"

and

$B =$ "the number of tails is even"

(a) $Pr(A) = Pr(B)$
(b) $Pr(A) < Pr(B)$
(c) $Pr(A) > Pr(B)$

The answer is (a).

• My initial understanding was that, the numbers $1$ to $5$ consist of $1, 2, 3, 4, 5$. There are $3$ odd numbers and $2$ even numbers. So $Pr(A) > Pr(B)$.

## 4 Answers

Why multiply by 100?
If you get the first one right, there are $3^{99}$ ways to get all the other problems wrong. But you could also only get the second problem right, with $3^{99}$ ways of getting all the other problems wrong, or only the third right. . . on to only getting the 100th question right with $3^{99}$ ways of getting numbers 1-99 wrong. So there are $100$ ways of getting one problem right times $3^{99}$ ways of getting all the rest wrong.

The second problem
If you flip a fair coin 5 times, then you could get 0, 1, 2, 3, 4, or 5 heads. Now there are 3 even and 3 odd head counts.

• Thank you! While all the other answers also made sense, this answer was easier to understand and more inuitive. – udpcon Feb 22 '18 at 21:25

For (a): since the answer is selected uniformly at random, the probability of $one$ correct answer out of 4 is $p=\frac{1}{4}, q=1-p= \frac{3}{4}$. Now, you have to select $exactly$ one out of 100, there are $\binom{100}{1}$ ways to do it.

Now, to get the probability of exactly one correct answer out 100 with this setup is $\binom{100}{1}pq^{99}$, hence the solution.

Answer to your first question

Why multiply by $100$.

Because, there are $100$ ways to select an answer that's correct.

Answer to your second question

Why isn't it equal.

Because if number of heads are odd, and since total flips are odd (5) too, the number of tails will be even automatically. So Event A is exactly the same as Event B.

For the first one you have that the probability of getting the correct answer is $\displaystyle\frac14$ and of getting the wrong answer is $\displaystyle\frac34$. Anyway the correct answer can be the one, the second, ecc. The probability is:

$$\binom{100}{1}\left(\frac 14\right)\left(\frac 34\right)^{99}$$

where $\displaystyle\binom{100}{1}$ is the binomial coefficient (The number of subset of 1 element of a set of 100 element)

For the second one you could notice that both head and tail have the same probability. Saying that the number of head is even is equal to say that the number of tail is odd (and that is equal to say that the number of head is odd)