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March 31, 2015

March 31, 2015

Posted by **Jay** on Friday, July 20, 2012 at 1:32am.

--4 From one suit, 3 from another

The probability of the event occurring is also acceptable. But, the answer is 99/16722971 (as far as the odds go

- Pre-Calc, Probability -
**Reiny**, Friday, July 20, 2012 at 9:18amCould be:

HHHHCCC

HHHHSSS

HHHHDDD

Now, each of those can be arranged in 7!/(4!3!) or 35 ways

let' calculate the prob of**one**of these, HHHHCCC

Prob(HHHHCCC)

= (13/52)(12/51)(11/50)(10/49)(13/48)(12/47)(11/46)

But the same prob is true for all of the 35 cases,

so multiply that by 35 to get

prob(4from one suit, 3 from another) =**1573/1029112**

so the prob of NOT any of those = 1 - 1573/1029112

= 1027539/1029112

So the odds in favour of 4 from one suit, 3 from another

= 1573 : 1027539

check my arithmetic, I did not get the same answer you gave.

- ARGGHH - error -Pre-Calc, Probability -
**Reiny**, Friday, July 20, 2012 at 9:36amoops, for some reason I read it as

4 Hearts, 3 from another

so it could be

HHHHCCC

HHHHSSS

HHHHDDD

DDDDHHH

DDDDCCC

DDDDSSS

SSSSHHH

SSSSCCC

SSSSDDD

CCCCHHH

CCCCSSS

CCCCDDD

So there are 12 of these and each can be arranged in 35 ways

so to get the prob of one of these 420 cases, multiply

(13/52)(12/51)(11/50)(10/49)(13/48)(12/47)(11/46) by 420

(my calculator is overheating)

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