Johnny the gambler tosses 6 fair dice. In order to win the jackpot, he

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

Hi anandch1994, I would like to answer your question.

The calculation comes from the formula to calculate the number of different ways we can arrange items, in which several items are similar to one another.
If you arrange 6 items, in which there are 3 similar items (A B C D D D), the number of ways: 6!/3!
If you arrange 6 items, in which there are 2 groups of 3 similar items (B B B G G G), the number of ways: 6!/(3!3!)

Hope this helps!

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
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Hi ScottTargetTestPrep

Could you please help me understand why do we divide by 3! two times in 6!/3!3!. Only G value is same as it can be 555 or 666 and the B value can be any number from 1 to 4. So why not 6!/3!?

6c3 x (2/6)^3 x (4/6)^3
One line solution if anyone want

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The number of ways to arrange B - B - B - G - G - G is given by a formula called "Permutation with repetition formula", also known as "Permutation of indistinguishable objects formula".

In our case, we have three indistinguishable letters B and three indistinguishable letters G, for a total of 6 letters. The 6! in the numerator comes from the total number of objects to be permuted and each 3! in the denominator comes from each of the three-letter groups.
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Hi Scott, are you missing the case 555666?

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