Click here to go to the

By Mooses41 (Mooses41) on Tuesday, June 08, 2004 - 01:08 am: Edit |

hey everyone,im having trouble with this one math problem, could someone please show me how to do it?

13 cards are delt from a standard deck. whats the probability of getting at least 3 aces? leave your answer in terms of combinations

thanks so much for the help

By Homer2568 (Homer2568) on Tuesday, June 08, 2004 - 10:53 am: Edit |

1-(4C2)(48C11)+(4C1)(48C12)+(48C13) all divided by (52C13)....this is the complement rule...so instead of figuring out the probability of getting at least 3 aces...at least 4...5...6...so on you do 1 minus the probability of getting 2 or 1 or 0 aces that way it makes it shorter and it workes nicer.

By Vanessa1378 (Vanessa1378) on Tuesday, June 08, 2004 - 01:04 pm: Edit |

Wouldn't it just be (4C4)(48C9)+(4C3)(48C10) since there are only 4 aces.

By Xiggi (Xiggi) on Tuesday, June 08, 2004 - 01:56 pm: Edit |

I would answer

P(exactly three aces) = C(4,3) * C(48,10) / C(52,13) = 0.04120048 or 0.04

Posting is currently disabled in this topic. Contact your discussion moderator for more information. |

Administrator's Control Panel -- Board Moderators Only Administer Page |