Posted by **Punkie** on Saturday, November 28, 2009 at 9:26pm.

I was wondering if someone could help me check my work on this problem and see if I'm correct.

Consider the experiment of drawing two cards without replacement from an ordinary deck of 52 playing cards.

What are the odds in favor of drawing a spade and a heart?

There are 52 playing cards in a deck

There are 13 spades and 13 hearts in deck

13 + 13 = 26

P(spade and heart) = 26/52

P(not a spade or heart) = 26/52

Odds in favor of A = Number of ways that A could occur/Number of ways that A could not occur

Odds in favor of A = 26/52 / 1- 26/52 = 0

- Math -
**Punkie**, Saturday, November 28, 2009 at 9:34pm
Or would it be

Odds in favor of A = 26/52 / 1 - 26/52 = 26 / 51

- Math -
**MathMate**, Saturday, November 28, 2009 at 9:37pm
"Odds in favor of A = 26/52 / (1- 26/52)= 0"

A in the above expression actually evaluates to 26/52 / (1-26/52) = (1/2) / (1/2) =1:1

However, I would reason it this way:

The first card can be either a spade or a heart, so the probability is 26/52.

The second card has to the complementary suit, namely a spade if the first one was a heart, and vice versa.

Probability of the second card is therefore limited to one suit out of 51 cards, or 13/51.

Probability of both events happening is

(26/52)*(13/51)=(1/2)*(13/51) = 13/102

Odds are

(13/102) / (1-13/102)

= 13/89

13:89

- Math -
**Punkie**, Saturday, November 28, 2009 at 10:03pm
I have a question about the answer, how did you get 13/89, I don't understand how you got 89

- Math -
**MathMate**, Saturday, November 28, 2009 at 11:16pm
Here it is, in a little more detail:

(13/102) / (1-13/102)

= (13/102) / ((102-13)/102)

=(13/102) / (89/102)

=13/89

=13:89

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