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# Combinations problem!

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Intern
Joined: 28 Aug 2011
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29 Aug 2011, 16:50
Why isn't the probability of getting 2 jacks, 2 queens, and 1 other card from 5 cards (4c2x4c2x48c1)/(52c5) not the same as 4/52(the chance for the first jack)x4/51(the chance for the second jack)x4/50(the chance for the first queen)x4/49(the chance for the second queen)x48/48 the chance for the other card? And how would you express the chance of getting a 4, 5, 6 in combinations(4/52x4/51x4/50), why wouldn't it be 4c1x4c1x4c1/52c3 in combinations? I made these problems up and can't seem to figure them out, gahhhhhhhh!!!

Look: the probability of getting 3 jacks and 2 kings in combinations is 4c3x4c2/52c5 which is 9.2344..., it's the same as 4/52x3/51x2/50x4/49x3/48, why can't the other ones work?

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Re: Combinations problem! [#permalink]

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31 Aug 2011, 22:16
JohnnyV wrote:
Why isn't the probability of getting 2 jacks, 2 queens, and 1 other card from 5 cards (4c2x4c2x48c1)/(52c5) not the same as 4/52(the chance for the first jack)x4/51(the chance for the second jack)x4/50(the chance for the first queen)x4/49(the chance for the second queen)x48/48 the chance for the other card? And how would you express the chance of getting a 4, 5, 6 in combinations(4/52x4/51x4/50), why wouldn't it be 4c1x4c1x4c1/52c3 in combinations? I made these problems up and can't seem to figure them out, gahhhhhhhh!!!

Probability of getting 2 jacks, 2 queens and 1 other card is 4C2 * 4C2 * 44C1 / 52C5
(Assuming that the other card cannot be a jack or a queen)
= 4*3 * 4*3 * 5*4*3*2* 44/ 2*2 * 52*51*50*49*48 = 4*3*4*3*(5*3*2)*44/ (52*51*50*49*48)

It is not the same as 4/52(the chance for the first jack) * 3/51(the chance for the second jack) * 4/50(the chance for the first queen) * 3/49(the chance for the second queen) * 44/48 the chance for the other card
Here, you are arranging the cards too. This is the probability of getting a jack first, the another jack, then a queen, then another queen and then any other card i.e. this is the probability of getting JJQQA
But you could get 2 jacks, 2 queens and another card in many other ways e.g. JQJQA or AQQJJ etc
When you find the probability this way, there is an implied sequence of events. To cover all the sequences that give you 2 jacks, 2 queens and another card, you need to multiply this probability by the total number of such sequences.
In how many ways can you arrange 2 jacks, 2 queens and another card? In 5!/(2!*2!) = 5*3*2
So total probability = (4/52) * (3/51) * (4/50) * (3/49) * (44/48) * 5*3*2 = 4*3*4*3*(5*3*2)*44/ 52*51*50*49*48

You see that the probability is the same in both the cases.

JohnnyV wrote:
Look: the probability of getting 3 jacks and 2 kings in combinations is 4c3x4c2/52c5 which is 9.2344..., it's the same as 4/52x3/51x2/50x4/49x3/48, why can't the other ones work?

Same concept here. Try and work it out.
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Kudos [?]: 17408 [0], given: 232

Re: Combinations problem!   [#permalink] 31 Aug 2011, 22:16
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# Combinations problem!

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