# Probability question explaination

• Apr 24th 2011, 09:03 AM
IanCarney
Probability question explaination
A standard deck of cards has had all the face cards (jacks, queens, and kings) removed so that only the ace (given a value of 1) through ten of each suit remains. A game is played in which two cards are drawn (without replacement) from this deck and a six-sided die is rolled. Find the probability that the sum of the cards and the die is 7.

This is the solution in the book:
http://img580.imageshack.us/img580/8245/datamang.png

I'm not sure if there's an easier way, but that has been a little confused. For starters, why is the (3,3,1) separated from the top set? Also, shouldn't the first part be (16/40)(16/39)(1/6)?

The way I attempted to set it up was like this:
(1,5,1)(2,4,1)(3,3,1)(4,2,1)(5,1,1)
+ (1,4,2)(2,3,2)(3,2,2)(4,1,2)
+ (1,3,3)(2,2,3)(3,1,3)
+ (1,2,4)(2,1,4)
+ (1,1,5)

Can anyone clarify the book solution? Why is it being done like that opposed to what I had done? Thanks a lot for the help!
• Apr 24th 2011, 09:42 AM
Plato
I would explain their solution this way.
The outcomes {1,1,5},{2,2,3}, & {3,3,1} all have the same probability
http://quicklatex.com/cache3/ql_3b0d...efd9f89_l3.png.

While every other 12 outcome each has probability http://quicklatex.com/cache3/ql_bdc8...84e47b4_l3.png.
• Apr 24th 2011, 09:46 AM
IanCarney
Quote:

Originally Posted by Plato
I would explain their solution this way.
The outcomes {1,1,5},{2,2,3}, & {3,3,1} all have the same probability
http://quicklatex.com/cache3/ql_3b0d...efd9f89_l3.png.

While every other 12 outcome each has probability http://quicklatex.com/cache3/ql_bdc8...84e47b4_l3.png.

Thanks but I'm still not sure why the first one for example is (16/40)(4/39)? Shouldn't it either be 4/40 or 16/39?
• Apr 24th 2011, 09:53 AM
Plato
Quote:

Originally Posted by IanCarney
Thanks but I'm still not sure why the first one for example is (16/40)(4/39)? Shouldn't it either be 4/40 or 16/39?

I tried to make clear that I find the book's solution confusing.
There are four outcomes listed on the first line, each has a probability of http://quicklatex.com/cache3/ql_bdc8...84e47b4_l3.png.
Multiply that by 4 you get the 16.