What's the Probability of Finding Cake in the Last Box?

[GlobalDuty]

you open box after box until you get the cake.
what's the probability that after opening 4 empty boxes, you open the last box with cake?

i know that the probability of finding cake in the first box is 1/5.
i think for the 5th box you multiply: (1/5)(1/4)(1/3)(1/2) to get the answer. is this right?

 

[CRGreathouse]

i know that the probability of finding cake in the first box is 1/5.

But you don't want the probability of cake, you want the probability of no cake.

Depending on how you interpret the question there are two (or more?) possible answers.

 

[mathman]

you open box after box until you get the cake.
what's the probability that after opening 4 empty boxes, you open the last box with cake?

i know that the probability of finding cake in the first box is 1/5.
i think for the 5th box you multiply: (1/5)(1/4)(1/3)(1/2) to get the answer. is this right?

This looks like a trick question. Simply put, if you have already opened 4 boxes and no cake, then the fifth box MUST have the cake (unless someone was fooling around and there was no cake at all). Therefore the probability is 1.

 

[GlobalDuty]

Depending on how you interpret the question there are two (or more?) possible answers.

yeah, this is bothering me

This looks like a trick question. Simply put, if you have already opened 4 boxes and no cake, then the fifth box MUST have the cake (unless someone was fooling around and there was no cake at all). Therefore the probability is 1.

the wording is bad, but its not a trick question. i guess better wording would be:
what's the probability that you open all the boxes and then find cake in the last one?

 

[CRGreathouse]

yeah, this is bothering me

If we assume what you write in the following paragraph, there is no ambiguity.

 

[HallsofIvy]

Here is the easy way to do it: by deciding which box to open first, which second, etc., you are specifying which box you will open last. Since the cake is equally likely to be in any of the 5 boxes, the probability that the cake is in that particular box is 1/5.


If you are not convinced that that answers the question you posed, then you can do it the hard way: the probability of NOT finding the cake in the first box is 4/5, the probability of NOT finding the cake in the second box (given that it was not in the first box) is 3/4, the probability of NOT finding the cake in the third box (given that it was not in the first or second boxes) is 2/3, and the probability of NOT finding the cake in the second box (given that it was not in the first, second, or third boxes) is 1/2.

The probability that the cake is in the fifth box is exactly the probability that it is NOT in the first box and NOT in the second box and NOT in the third box and NOT in the fourth box: (4/5)(3/4)(2/3)(1/2)= 1/5.

That, of course, is just the probability that it was in that particular box a-priori.

 

[GlobalDuty]

o ok, thanks for the responses.
the problem was bad to begin with.