# Thread: There are 100 photos in the envelope, among whose is the wanted one?

1. ## There are 100 photos in the envelope, among whose is the wanted one?

There are There are 100 photos in the envelope, among whose is the wanted one. Tim picks up 10 photos randomly. What is the probability that the wanted one is among these ten photos?

My Solution: I would say, since we have ONLY 1 wanted photo and we pick 10 photos therefore the probability that the photo Tim wants is one of those 10 is
C(1,100) / C(10,100)

My friend says : No! It's (C(1,1) * C(9,99)) / C(10,100)

which one is correct and WHY? Thank you

2. If we group the 100 in groups of 10, each group has a 10% chance so 10% is the probability that 10 contains the 1.

3. Therefore, non of mine and my friend's solutions are correct?

4. Hello, Narek!

Therefore, neither mine and my friend's solutions are correct?

Your friend's solution, which should be written: $\dfrac{C(1,1)\cdot C(99,9)}{C(100,10)}$
. . is equal to $\dfrac{1}{10}$ . . . and is correct!

His/her reasoning is correct.

There are $C(100,10)$ possible outcomes; that's the denominator.

There is 1 special photo and 99 Others.

There is $C(1,1)$ way to get the special photo.
And there are $C(99,9)$ ways to choose 9 Others.

Hence, there are: . $C(1,1)\cdot C(99,9)$ ways to select 10 photos,
. . including the special one; that's the numerator.

Please pass this on to your friend: . Good work!

5. Originally Posted by Narek
My Solution: I would say, since we have ONLY 1 wanted photo and we pick 10 photos therefore the probability that the photo Tim wants is one of those 10 is C(1,100) / C(10,100)
In addition to what has been written already, maybe the following helps you seeing why this is wrong:

Suppose there are only 10 or 11 photos instead of 100 (and again, there's 1 you want, and you pick 10). Then according to your method, what are the probabilities of the right photo being in your pick?

(condsidering that with 10 photos the probability should be 1, and with 11 it should be less than 1, obviously)