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Probability 1

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Probability 1

by karthikpandian19 » Mon Dec 26, 2011 6:45 pm
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×(2^2/3^6)
10×(2^3/3^6)
20×(2^3/3^5)
20×(2^3/3^6)
20×(2^2/3^5)
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by Anurag@Gurome » Mon Dec 26, 2011 7:36 pm
karthikpandian19 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×(2^2/3^6)
10×(2^3/3^6)
20×(2^3/3^5)
20×(2^3/3^6)
20×(2^2/3^5)
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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by ronnie1985 » Mon Dec 26, 2011 7:49 pm
P(5 or 6) = 1/3
It is binomial probability, q = 2/3
P(Win) = 6C3(1/3)^3*(2/3)^3
P = 20*2^3/3^6
(C) is correct
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by karthikpandian19 » Tue Dec 27, 2011 9:53 pm
OA is D
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by karthikpandian19 » Tue Dec 27, 2011 9:58 pm
Anurag,

Can you explain the process of this question?
step by step

I am weak in Probability & Combinations
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