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Updated on: 29 Dec 2017, 01:06
Originally posted by gmatter0913 on 05 Aug 2013, 22:47.
Last edited by Bunuel on 29 Dec 2017, 01:06, edited 2 times in total.
Last edited by Bunuel on 29 Dec 2017, 01:06, edited 2 times in total.
Edited the question, renamed the topic.
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Bill has a set of 6 black cards and a set of 6 red cards. Each card has a number from 1 through 6, such that each of the numbers 1 through 6 appears on 1 black card and 1 red card. Bill likes to play a game in which he shuffles all 12 cards, turns over 4 cards, and looks for pairs of cards that have the same value. What is the chance that Bill finds at least one pair of cards that have the same value?
(A) \(\frac{8}{33}\)
(B) \(\frac{62}{65}\)
(C) \(\frac{17}{33}\)
(D) \(\frac{103}{165}\)
(E) \(\frac{25}{33}\)
(A) \(\frac{8}{33}\)
(B) \(\frac{62}{65}\)
(C) \(\frac{17}{33}\)
(D) \(\frac{103}{165}\)
(E) \(\frac{25}{33}\)
06 Aug 2013, 00:23
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Probability at least one pair equal \(= 1-P(none)\).
Now I am gonna find the probability of picking all different values.
You can chose the first card is 12 ways ( every card is OK ), the second in 10 ways (there are 11 cards remaining, and 1 has the same value as the first one), the third in 8 ways (there are 10 cards remaining, 2 with the same value as the first ones), and the fourth in 6 ways (same reasoning).
The first is chosen out of a pool of 12(all cards), the second out of a pool of 11, the third out of a pool of 10, the fourth out of a pool of 9.
\(P=1-\frac{12}{12}*\frac{10}{11}*\frac{8}{10}*\frac{6}{9}=1-\frac{16}{33}=\frac{17}{33}.\)
Hope it's clear
Now I am gonna find the probability of picking all different values.
You can chose the first card is 12 ways ( every card is OK ), the second in 10 ways (there are 11 cards remaining, and 1 has the same value as the first one), the third in 8 ways (there are 10 cards remaining, 2 with the same value as the first ones), and the fourth in 6 ways (same reasoning).
The first is chosen out of a pool of 12(all cards), the second out of a pool of 11, the third out of a pool of 10, the fourth out of a pool of 9.
\(P=1-\frac{12}{12}*\frac{10}{11}*\frac{8}{10}*\frac{6}{9}=1-\frac{16}{33}=\frac{17}{33}.\)
Hope it's clear
06 Aug 2013, 05:59
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gmatter0913
Your denominator is correct \(12C4=495\). The problem is the numerator: the probability of picking 4 different values among 6 possible values: \(6C4=15\). But because each of the four cards can be black or red, each of the four slots can have 2 colors , so we have to multiply 15 by \(2*2*2*2=16\) (each of the four cards can be B or R).
Hope I've explained myself well.
So now \(P=\frac{495-15*16}{495}=\frac{17}{33}\).
)
need some help here...ohh and btw i did not use the
