Posted by Katie on Friday, June 1, 2012 at 12:26am.
I'll solve #2 first, where the two cards requested are mutually exclusive.
Since there are two black jacks, and 4 aces, we have
P(BJ,A)=2/52*(4/51)=8/2652=2/663
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For problem #1, we denote the events as:
B=black card, B'=red card
F=face card, F'=non-face card
The possible successful outcomes can be split into two stages, first and second card, denoted by the couple ( , ).
First a face card then a black card:
P(B'F,B)=(6/52)(26/51) (red face,black)
P(BF,BF')=(6/52)(20/51) (black face, black non-face)
Note: (black face, black face) is included in the next case.
Then a black card followed by a face card:
(BF,F)=(6/52)(11/51) (black face,any face)
(BF',F)=(20/52)(12/51) (black non-face, any face)
All these add up to 97/442.
We can also calculate the unsuccessful outcomes and subtract from 1 to get the successful outcomes.
The unsuccessful outcomes are:
P(F',F')=(40/52)(39/52) (two non-face)
P(B',B')=(26/52)(25/52) (two reds)
P(BF,B'F')=(6/52)(20/52) (black face,red non-face)
P(B'F',BF)=(20/52)(6/52) (red non-face, black face)
Out of this, we have double-counted, and therefore should subtract
P(F'B',F'B')=(20/52)(19/51) (both cards red non face)
Summing them all up we get 345/442 for the probability of unsuccessful outcomes.
Subtract from 1 to get 97/442 as the probability of successful outcomes. This is the same probability as we calculated above.