problem solving - combinations + probability

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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?

20*(2^2/2^3)

10*(2^3/2^6)

20*(2^3/2^5)

20*(2^3/2^6)

20*(2^2/2^5)

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by Anurag@Gurome » Wed Sep 26, 2012 7:45 pm
arellalu 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?

20*(2^2/2^3)

10*(2^3/2^6)

20*(2^3/2^5)

20*(2^3/2^6)

20*(2^2/2^5)
Check the answer choices please, it should be "3" instead of "2" in the denominator.

Probability of getting a 5 = 1/6
Probability of getting a 6 = 1/6
Probability of getting a 5 or 6 = 1/6 + 1/6 = 1/3
Now, probability of getting 1, 2, 3, or 4 = 1 - probability of getting a 5 or 6 = 1 - 1/3 = 2/3
Probability of getting exactly 3 times a result of 5 or 6 = 1/3 * 1/3 * 1/3 * 2/3 * 2/3 * 2/3 = 2^3/3^6
Johny can get a 5 or 6 in any order in 6 times in 6C3 = 20 ways

Therefore, required no. of ways = [spoiler]20 * 2^3/3^6 [/spoiler]

The correct answer is D.
Anurag Mairal, Ph.D., MBA
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