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12 Nov 2019, 02:00
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D
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Difficulty:
55%
(hard)
Question Stats:
70% (02:21) correct
30%
(02:56)
wrong
based on 358
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Johnny the gambler tosses 6 fair dice. In order to win the jackpot, he needs a 5 or 6 on exactly three of the dice. What is the probability that Johnny will win?
A. \(20*(\frac{2^2}{3^6})\)
B. \(10*(\frac{2^3}{3^6})\)
C. \(20*(\frac{2^3}{3^5})\)
D. \(20*(\frac{2^3}{3^6})\)
E. \(20*(\frac{2^2}{3^5})\)
Are You Up For the Challenge: 700 Level Questions
A. \(20*(\frac{2^2}{3^6})\)
B. \(10*(\frac{2^3}{3^6})\)
C. \(20*(\frac{2^3}{3^5})\)
D. \(20*(\frac{2^3}{3^6})\)
E. \(20*(\frac{2^2}{3^5})\)
Are You Up For the Challenge: 700 Level Questions
Johnny the gambler tosses 6 plain dice. In order to win the jackpot, he needs a 5 or 6 on exactly three of the dice. What are Johnny's chances to win?
12 Nov 2019, 03:21
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P of 5 or 6 = 2/6 ; 1/3
and P of not 5 or 6 ; 4/6 ; 2/3
total P of getting in exactly 3 chances ; 6c3 * (1/3)^3* ( 2/3)^3 =
\(20*(\frac{2^3}{3^6})\)
IMO D
and P of not 5 or 6 ; 4/6 ; 2/3
total P of getting in exactly 3 chances ; 6c3 * (1/3)^3* ( 2/3)^3 =
\(20*(\frac{2^3}{3^6})\)
IMO D
Bunuel
20 Nov 2019, 19:27
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Bunuel
Let G denote the event that the outcome is 5 or 6, and let B denote the event that the outcome is 1, 2, 3 or 4. Then, P(G) = 2/6 = 1/3 and P(B) = 4/6 = 2/3.
Notice that the probability of G-G-G-B-B-B is (1/3)^3 * (2/3)^3 = (2^3)/(3^6). The condition of “exactly 3 of the 6 tosses is 5 or 6” can also be satisfied in other ways, such as B-B-B-G-G-G or B-G-B-G-B-G. Each of those other ways also has a probability of (2^3)/(3^6). The total number of different ways is the number of arrangements of G-G-G-B-B-B; which is 6!/(3!*3!) = (6 x 5 x 4)/3! = 5 x 4 = 20.
Thus, the probability of obtaining 5 or 6 in exactly 3 of the 6 rolls is 20 * (2^3)/(3^6).
Answer: D
