# Conditional probability question help

• Nov 8th 2010, 09:40 PM
demo12
Conditional probability question help
Can't seem to figure this question out. Can someone please show me how to approach it?

At bridge hand (13 cards) contains seven clubs, four hearts and two spades. Two cards are drawn at random from the hand without replacement. Find:

Pr [One card is a club | at least one card is a heart]

Thanks.
• Nov 8th 2010, 11:11 PM
Unknown008
Use the equation form Baye's Theorem.

$P(\text{One card is a club | at least one card is a heart}) = \dfrac{P(\text{One card is a club} \cap \text{at least one card is a heart})}{P(\text{at least one card is a heart})}$
• Nov 9th 2010, 01:13 AM
demo12
Quote:

Originally Posted by Unknown008
Use the equation form Baye's Theorem.

$P(\text{One card is a club | at least one card is a heart}) = \dfrac{P(\text{One card is a club} \cap \text{at least one card is a heart})}{P(\text{at least one card is a heart})}$

Using that, I need getting 7/12 as my answer which is incorrect. The back of the book shows 2/3. Do you mind telling me which numbers I'm supposed to plug in?

Thanks
• Nov 9th 2010, 01:38 AM
Unknown008
P(at least one card is a heart) = (P(1st card heart) x P(2nd card not heart)) + (P(1st card not heart) x P(2nd card heart)) + (P(both cards heart)) =

$\left(\dfrac{4}{13} \times \dfrac{9}{12}\right) + \left(\dfrac{9}{13} \times \dfrac{4}{12}\right) + \left(\dfrac{4}{13} \times \dfrac{3}{12}\right) = \dfrac{7}{13}$

For the numerator, you must have one club and a heart, or a heart and a club.

P(Heart and club) = 4/13 x 7/12 = 7/39

P(club and heart) = 7/13 x 4/12 = 7/39

Hence, we get: $\dfrac{\frac{14}{39}}{\frac{7}{13}} = \dfrac23$

Another way of looking at it is that you have the following combinatios:

You want to have:
HC

out of:
HH
SH
CH

(H - heart, C- club, S - spade)

This gives a total of $\dfrac{^4C_1 \times ^7C_1}{^4C_2 + (^2C_1 \times ^4C_1) + (^7C_1 \times ^4C_1)}= \dfrac23$
• Nov 9th 2010, 10:08 AM
demo12
Thanks! I missed a couple steps in my calculations.