Probability IB Math HL: Is it a difficult question to solve?

[coolfish21]

TL;DR

I have not been able to figure out the solution. Can anyone help me with this?

A person picks 8 cards without replacement from a bag containing cards numbered from 1 to 6 (117 in total). What is the probability that the sum of those 8 cards is 28? Given that P(1)=8/117, P(2)=14/117, P(3)=34/117, P(4)=39/117. P(5)=14/117 and P(6)=8/117.

A person picks 8 cards without replacement from a bag containing cards numbered from 1 to 6 (117 in total). What is the probability that the sum of those 8 cards is 28? Given that P(1)=8/117, P(2)=14/117, P(3)=34/117, P(4)=39/117. P(5)=14/117 and P(6)=8/117.

 

[mfb]

Where does this problem come from?
What did you try so far?
This will probably need ugly casework. With replacement it would be much easier.

 

[PeroK]

Summary:: I have not been able to figure out the solution. Can anyone help me with this?

A person picks 8 cards without replacement from a bag containing cards numbered from 1 to 6 (117 in total). What is the probability that the sum of those 8 cards is 28? Given that P(1)=8/117, P(2)=14/117, P(3)=34/117, P(4)=39/117. P(5)=14/117 and P(6)=8/117.

A person picks 8 cards without replacement from a bag containing cards numbered from 1 to 6 (117 in total). What is the probability that the sum of those 8 cards is 28? Given that P(1)=8/117, P(2)=14/117, P(3)=34/117, P(4)=39/117. P(5)=14/117 and P(6)=8/117.

I suggest you write a computer program to do it.

 

[mathman]

First step: list all combination which add to 28. The probability in each case is straightforward. Then add.