It's ridiculous to get it right at the last guess

Probability Level 2

A student was sleeping in class,so the teacher called him to answer a question which has four choices,(A),(B),(C),(D).

The student was sleeping,so he did not know what to answer.

He randomly guessed one of the four options,and got it wrong.

He guessed again with another option,but still got it wrong.

He finally got it right at the fourth time(of course he would not guess more than four times)

What is the probability of finally getting it right at the fourth time if there are four options?

\frac1{256}

\frac1{24}

\frac16

\frac14

\frac1{16}

2 solutions

Romain Bouchard
Jul 31, 2018

To get it wrong the three first times, the student needs to get it wrong the first time ( $3$ chances out of $4$ ) AND the second time ( $2$ chances out of the $3$ remaining answers) AND the third time ( $1$ chance out of the $2$ remaining answer) hence the probability is $P=\frac{3}{4}\times\frac{2}{3}\times\frac{1}{2} = \boxed{\frac{1}{4}}$ .

What's neat is that for $n$ possible answers his probability to get it right at the last possible time is $\frac{1}{n}$ .

Good solution!

P.S.It's strange that I asked some people this,and most of them answered 1/24.This is why I posted this problem.

X X - 2 years, 10 months ago

K T
Nov 9, 2020

Consider any order the student might give 4 answers. Because there are 4 possible answers, by symmetry, the probability that the correct answer happens to be the last one, is of course 1/4.

It's ridiculous to get it right at the last guess

2 solutions

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