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Math Expert V
Joined: 02 Sep 2009
Posts: 59677
Johnny the gambler tosses 6 fair dice. In order to win the jackpot, he  [#permalink]

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Difficulty:   65% (hard)

Question Stats: 60% (02:34) correct 40% (02:44) wrong based on 67 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

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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Re: Johnny the gambler tosses 6 fair dice. In order to win the jackpot, he  [#permalink]

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1
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 wrote:
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

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?
Target Test Prep Representative V
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Re: Johnny the gambler tosses 6 fair dice. In order to win the jackpot, he  [#permalink]

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Bunuel wrote:
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

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).

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Joined: 02 Oct 2016
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Johnny the gambler tosses 6 fair dice. In order to win the jackpot, he  [#permalink]

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ScottTargetTestPrep wrote:
Bunuel wrote:
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

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).

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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Re: Johnny the gambler tosses 6 fair dice. In order to win the jackpot, he  [#permalink]

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anandch1994 wrote:
ScottTargetTestPrep wrote:
Bunuel wrote:
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

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).

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!?

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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Re: Johnny the gambler tosses 6 fair dice. In order to win the jackpot, he  [#permalink]

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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  [#permalink]

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anandch1994 wrote:

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!?

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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# Scott Woodbury-Stewart

Founder and CEO

Scott@TargetTestPrep.com

See why Target Test Prep is the top rated GMAT quant course on GMAT Club. Read Our Reviews

If you find one of my posts helpful, please take a moment to click on the "Kudos" button. Johnny the gambler tosses 6 fair dice. In order to win the jackpot, he   [#permalink] 04 Dec 2019, 18:57
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# Johnny the gambler tosses 6 fair dice. In order to win the jackpot, he  