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

• Nov 29th 2010, 04:36 AM
Narek
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
• Nov 29th 2010, 04:39 AM
dwsmith
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.
• Nov 29th 2010, 04:44 AM
Narek
Therefore, non of mine and my friend's solutions are correct?
• Nov 29th 2010, 12:10 PM
Soroban
Hello, Narek!

Quote:

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.