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Bunuel
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Johnny the gambler tosses 6 fair dice. In order to win the jackpot, he 12 Nov 2019, 02:00
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71% (02:21) correct 29% (02:54) wrong based on 387 sessions

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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

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

Bunuel
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

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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?
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Johnny the gambler tosses 6 fair dice. In order to win the jackpot, he 20 Nov 2019, 19:27
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Bunuel
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

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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?
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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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Johnny the gambler tosses 6 fair dice. In order to win the jackpot, he 27 Nov 2019, 19:57
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Bunuel
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

Show Spoiler
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?

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!?
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Johnny the gambler tosses 6 fair dice. In order to win the jackpot, he 28 Nov 2019, 03:28
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anandch1994
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Bunuel
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

Show Spoiler
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?

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

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!?
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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!
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Johnny the gambler tosses 6 fair dice. In order to win the jackpot, he 28 Nov 2019, 03:57
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6c3 x (2/6)^3 x (4/6)^3
One line solution if anyone want

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Johnny the gambler tosses 6 fair dice. In order to win the jackpot, he 04 Dec 2019, 18:57
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anandch1994


Hi ScottTargetTestPrep

Could you please help me understand why do we divide by 3! two times in 6!/3!3!. Only G value is the 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!?
Show more

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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Johnny the gambler tosses 6 fair dice. In order to win the jackpot, he 17 Mar 2024, 07:01
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Hi Scott, are you missing the case 555666?
ScottTargetTestPrep

Bunuel
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

Show Spoiler
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?
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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Johnny the gambler tosses 6 fair dice. In order to win the jackpot, he 07 Apr 2026, 19:28
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How I solved:
For probability questions like these, it's often easy to structure the problem by setting up a fraction, which is total number of desired outcomes over total number of possible outcomes.

The total number of possible outcomes is usually the simpler one. In this case there are 6 possible outcomes for each dice roll of which there are 6, so its 6^6.

For total number of desired outcomes, we need to take all of the cases where 3 out of the 6 dice rolls are either a 5 or a 6 and the others are not.

First, consider the total number of ways to pick 3 from 6: 6C3. Next, for each of the three slots which have our desired outcome, there are two choices they can be: either a 5 or a 6. That's 2 * 2 * 2. For the last three slots, they must not be 5 or 6, so they must be 1,2,3 or 4. Therefore we multiply by 4 * 4 * 4 to account for each of these possible outcomes.

Altogether, it is 6C3*2*2*2*4*4*4/6^6

Simplified: 20* 2^9/2^6*3^6

Cancel out: 20* 2^3/3^6
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