# Probability

• Sep 20th 2007, 02:29 AM
r7iris
Probability
An instructor gives her class a set of 10 porblems with the information that the final exam will consist of a random selection of 5 of them. If a student has figured out how to do 7 of the the problems, what is the probability that the student will answer correctly
a) all 5 problems?
b) at least 4 of the problems?
• Sep 20th 2007, 08:40 AM
ThePerfectHacker
Quote:

Originally Posted by r7iris
An instructor gives her class a set of 10 porblems with the information that the final exam will consist of a random selection of 5 of them. If a student has figured out how to do 7 of the the problems, what is the probability that the student will answer correctly
a) all 5 problems?
b) at least 4 of the problems?

Use the formula ${n\choose m}p^m (1 - p)^{n-m}$.
Where $p=7/10 = .7$ because that is the probability that she knows.
• Sep 20th 2007, 04:09 PM
Soroban
Hello, r7iris!

I think I've worked it out . . .

Quote:

An instructor gives her class a set of 10 problems with the information
that the final exam will consist of a random selection of 5 of them.

If a student has figured out how to do 7 of the the problems,
what is the probability that the student will answer correctly
(a) all 5 problems?
(b) at least 4 of the problems?

(a) It doesn't matter which five problems the instructor picks.
We are concerned with whether the student's seven prepared solutions
. . include the five problems.

There are: . ${10\choose7} \,=\,120$ ways that the student can choose 7 problems.

To get all 5 problems correct, his 7 preparations must include:
. . the 5 problems chosen by the instructor (one way),
. . and any 2 of the other 5 problems: . ${5\choose2} \,=\,10$ ways.
Hence, he has: . $1\cdot10 \:=\:10$ ways to get all 5 problems correct.

Therefore: . $P(\text{5 correct}) \;=\;\frac{10}{120} \;=\;\frac{1}{12}$

(b) To get 4 or 5 problems correct . . .

We already know there are $10$ ways to get 5 correct.

To get 4 correct, he must choose:
. . 4 of the chosen 5 problems: . ${5\choose4} \,=\,5$ ways.
. . and 3 from the other five: . ${5\choose3} \,=\,10$ ways.
Hence, he has: . $5\cdot10 \,=\,50$ ways to get 4 correct.

Therefore, he has: . $10 + 50 \:=\:60$ ways to get 4 or 5 correct
. . and: . $P(\text{4 or 5 correct}) \;=\;\frac{60}{120} \:=\:\frac{1}{2}$