# Math Help - Prob. Question

1. ## Prob. Question

Question
5 boys (call them A,B,C,D,E) place their books in a bag. The 5 books are are drawn out in random order and given back to the 5 boys. What is the probability that exactly 3 boys will receive their original book?

Help wanted please. The only way I could think of to solve this problem is to list the number of ways 3 boys could receive their original book and sum the individual probabilities of each of them occurring. This is tedious and can't be sure if I have them all.
If someone could list an alternative method, it would be greatly appreciated.

2. Originally Posted by I-Think
Question
5 boys (call them A,B,C,D,E) place their books in a bag. The 5 books are are drawn out in random order and given back to the 5 boys. What is the probability that exactly 3 boys will receive their original book?

Help wanted please. The only way I could think of to solve this problem is to list the number of ways 3 boys could receive their original book and sum the individual probabilities of each of them occurring. This is tedious and can't be sure if I have them all.
If someone could list an alternative method, it would be greatly appreciated.
If exactly three boys have their original books then the other two books are switched. Do you see how this helps?

3. Hello, I-Think!

Five boys place their books in a bag.
The books are are drawn out in random order and given back to the boys.
What is the probability that exactly 3 boys will receive their original book?

The only way I could think of to solve this problem is to
list the number of ways 3 boys could receive their original books.
You don't have to list them . . .

Select three of the five boys who will get their own books.
. . There are: . $_5C_3 \:=\:{5\choose3} \:=\:10$ ways.

The other two boys have simply switched books: . $1$ way.

Hence, there are: . $10\cdot1 \:=\:10$ ways for 3 boys to get their own books.

There are: . $5! \,=\,120$ ways to return the books.

Therefore: . $P(\text{3 get own books}) \;=\;\dfrac{10}{120} \;=\;\dfrac{1}{12}$

4. Alternatively ask how many ways to select the two books that get switched, which is C(5,2) = 10, and you get the same answer as Soroban.