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Mathematics 10 Online
OpenStudy (lgbasallote):

Three cards are given to you from a well-shuffled deck of cards. What is the probability that you get all 3 cards from the same suit?

OpenStudy (anonymous):

thirteen choose three over fifty two choose three

OpenStudy (anonymous):

times four

OpenStudy (lgbasallote):

is this \[\frac 1{52 C3}\]

OpenStudy (kinggeorge):

Try the method I was talking about a little earlier.

OpenStudy (kinggeorge):

You first need to find the possible ways you can choose three cards from the same suit, and then divide by all the possible three card hands. So you have Construction: Choices: 1. Choose a suit \(\binom41\) and then 2. Choose a 3 card set from 13 \(\binom{13}{3}\) So there are \[\binom41\cdot\binom{13}{3}\]3card hands where all the cards are from the same suit.

OpenStudy (lgbasallote):

so the probability would be \[\frac 1{4 C 1 \times 13 C 3}\]

OpenStudy (kinggeorge):

Now we need to find all 3 card hands total. This is just \(\binom{52}{3}\) (it's a one step process with how I was doing it before). Thus, your probability would be \[\Large P=\frac{\binom{4}{1}\cdot\binom{13}{3}}{\binom{52}{3}}\]

OpenStudy (lgbasallote):

probability is.... the possible events over the total right?

OpenStudy (kinggeorge):

We have to divide by the total number of hands to find probability.

OpenStudy (kinggeorge):

Right.

OpenStudy (lgbasallote):

oh. i get the logic now

OpenStudy (lgbasallote):

thanks

OpenStudy (kinggeorge):

You're welcome.

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