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August 29, 2015
Posted by **Katie** on Friday, June 1, 2012 at 12:26am.

1) You choose two cards and one is a face card and the other is black

2) A black jack and then an ace are drawn

- ALGEBRA II -
**MathMate**, Friday, June 1, 2012 at 10:32amI'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.