• January 7th 2009, 05:44 PM
asugirl87
An instructor gives his class a set of 11 problems and announces that the next quiz will consist of a random selection of 5 of them. If a student has figured out how to answer 6 of the problems correctly, what is the probability the he or she will answer correctly:
(a) All 5 problems?
(b) At least 4 problems?
• January 7th 2009, 07:18 PM
Soroban
Hello, asugirl87!

Quote:

An instructor gives his class a set of 11 problems and announces
that the next quiz will consist of a random selection of 5 of them.
If a student has figured out how to answer 6 of the problems correctly,
what is the probability the he or she will answer correctly:
(a) All 5 problems?

There are: ${11\choose5} \,=\,462$ possible quizzes.

There are ${6\choose5} = 6$ ways that the five questions are among the six the student knows.

Therefore: . $P(\text{5 right}) \:=\:\frac{6}{462} \:=\:\frac{1}{77}$

Quote:

(b) At least 4 problems?

"At least four" means 4 or 5 right.

There are ${6\choose4} = 15$ ways that four questions are among the six the student knows.

There are 6 ways that all five questions are among the six the student knows.

Hence, there are: . $16 + 6 \,=\,21$ ways to get 4 or 5 right.

Therefore: . $P(\text{4 or 5 right}) \:=\:\frac{21}{462} \:=\:\frac{1}{22}$