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Johnny the gambler

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by smodak » Sun Jun 26, 2011 11:14 am
Johnny the gambler tosses 6 plain dice. In order to win the jackpot he has to receive exactly 3 times a result of 5 or 6. What are Johnny's chances to win?

OA:[spoiler]20×(2^3/3^6)[/spoiler]

Please explain how you arrived at the answer:

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by Frankenstein » Mon Jun 27, 2011 9:42 pm
Hi,
Out of 6 tosses, 3 should be (5 or 6) and the remaining 3 should be from (1,2,3,4)
Probability of getting 5 or 6 is p = 2/6 = 1/3
Probability of getting a number from(1,2,3,4) is q = 4/6 = 2/3
Probability of getting 3ps and 3qs in 6 throws is 6C3.p^3.q^3 = (6.5.4/6)*(1/3)^3*(2/3)^3
= 20*(2^3)/3^6
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by winniethepooh » Mon Jun 27, 2011 10:25 pm
Frankenstine have you considered 5, 5, and 6 OR 5, 6, and 6 OR 6, 6, and 5 as a favorable event?
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by winniethepooh » Mon Jun 27, 2011 10:25 pm
Frankenstine have you considered 5, 5, and 6 OR 5, 6, and 6 OR 6, 6, and 5 as a favorable event?
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by Frankenstein » Mon Jun 27, 2011 10:34 pm
winniethepooh wrote:Frankenstine have you considered 5, 5, and 6 OR 5, 6, and 6 OR 6, 6, and 5 as a favorable event?
5,5,5
5,6,6(3 permutations)
6,5,5(3 permutations)
6,6,6
These are the favorable events.
Cheers!

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by GMATGuruNY » Tue Jun 28, 2011 12:04 am
smodak wrote:Johnny the gambler tosses 6 plain dice. In order to win the jackpot he has to receive exactly 3 times a result of 5 or 6. What are Johnny's chances to win?

OA:[spoiler]20×(2^3/3^6)[/spoiler]

Please explain how you arrived at the answer:

Source: Master GMAT
P(exactly n times) = P(one way) * total possible ways.

Let G = 5 or 6
Let B = not 5 or 6

P(G) = 2/6 = 1/3
P(B) = 1-1/3 = 2/3.

P(one way):
P(GGGBBB) = (1/3)³(2/3)³ = 2³/3�.

Total possible ways:
Any arrangement of GGGBBB will yield exactly 3 G's.
Thus, the result above must be multiplied by the number of ways to arrange GGGBBB = 6!/3!3! = 20.

P(exactly 3 G's) = 20*(2³/3�).
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