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

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

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12 Nov 2019, 02:00
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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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12 Nov 2019, 03:21
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?
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Re: Johnny the gambler tosses 6 fair dice. In order to win the jackpot, he  [#permalink]

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20 Nov 2019, 19:27
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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Johnny the gambler tosses 6 fair dice. In order to win the jackpot, he  [#permalink]

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27 Nov 2019, 19:57
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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28 Nov 2019, 03:28
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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28 Nov 2019, 03:57
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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04 Dec 2019, 18:57
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.
_________________

# Scott Woodbury-Stewart

Founder and CEO

Scott@TargetTestPrep.com
122 Reviews

5-star rated online GMAT quant
self study course

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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