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Mathematics 9 Online
OpenStudy (anonymous):

Q: From a standard deck of 52 card, 2 cards are dealt. What is the probability that one is a 2 and one is a heart? How do I deal with the case where the 2 is a heart?

jimthompson5910 (jim_thompson5910):

I'd break it up into two cases: Case 1 being where you draw a 2 of hearts Case 2 being where you don't draw a 2 of hearts (so you draw a 2 of some other suit) Each case has the second card being a heart ------------------------------------------------------- Case 1: The first card is the two of hearts P(drawing a 2 and drawing a heart) = (1/52)*(12/51) = 12/2652 = 1/221 ------------------------------------------------------- Case 1: The first card is a 2, but it is not the two of hearts (so it's a 2 of spades, 2 of clubs, or 2 of diamonds) P(drawing a 2 and drawing a heart) = (3/52)*(13/51) = 39/2652 = 1/68

jimthompson5910 (jim_thompson5910):

Case 1 and Case 2 are mutually exclusive events. The two events cannot happen at the same time. Because they are mutually exclusive events, you can add the probabilities up to get your final answer.

OpenStudy (anonymous):

Oh, that makes sense. I didn't think of it as an "or" case. Thanks.

jimthompson5910 (jim_thompson5910):

oops typo, meant to say "case 2" and not repeat "case 1" but you get the idea

jimthompson5910 (jim_thompson5910):

you're welcome

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