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After each throw of a red die, the face that shows is marked

Expert replies
by BTGmoderatorDC » Thu Jul 26, 2018 10:14 pm
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After each throw of a red die, the face that shows is marked with a blue stripe. What is the probability that after 6 throws all faces of the die will be marked blue?

A. 3/126

B. 5/324

C. 4/216

D. 4/182

E. 5/216
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Source: — Problem Solving |

by Jay@ManhattanReview » Thu Jul 26, 2018 11:31 pm
BTGmoderatorDC wrote:After each throw of a red die, the face that shows is marked with a blue stripe. What is the probability that after 6 throws all faces of the die will be marked blue?

A. 3/126

B. 5/324

C. 4/216

D. 4/182

E. 5/216
1. First throw: Since all the six faces are red, any face would do; so the probability = 6/6 = 1
2. Second throw: Since one face is blue and five are red, and we need any of the five faces, the probability = 5/6
3. Third throw: Since the two faces are blue and four are red, and we need any of the four faces, the probability = 4/6
4. Fourth throw: Since the three faces are blue and three are red, and we need any of the three faces, the probability = 3/6
5. Fifth throw: Since the four faces are blue and two are red, and we need any of the two faces, the probability = 2/6
6. Sixth throw: Since the five faces are blue and only one is red, and we need that one red face, the probability = 1/6

Combined probability = 1*5/6*4/6*3/6*2/6*1/6 = 5/324

The correct answer: B

Hope this helps!

-Jay
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by Jeff@TargetTestPrep » Mon Jul 30, 2018 10:48 am
BTGmoderatorDC wrote:After each throw of a red die, the face that shows is marked with a blue stripe. What is the probability that after 6 throws all faces of the die will be marked blue?

A. 3/126

B. 5/324

C. 4/216

D. 4/182

E. 5/216
There is a 1/1 chance of rolling any number on the first roll, after which the side with that number is marked with blue. In the second roll, there is a 5/6 probability that a red side shows up; in the third roll, there is a 4/6 probability that a red side shows up, and so on.

The total probability is:

1 x 5/6 x 4/6 x 3/6 x 2/6 x 1/6

1 x 5/6 x 2/3 x 1/2 x 1/3 x 1/6 = 10/648 = 5/324

Answer: B

Jeffrey Miller
Head of GMAT Instruction
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by swerve » Tue Jul 31, 2018 10:17 am
Total outcomes = 6^6. (Cause any one of the six faces may appear on the top in each throw).

Favorable outcomes = 6*5*4*3*2*1 = 6! = 720 (Cause the top face shouldn't repeat hence there are 6 acceptable outcomes for the first throw and subsequently 5 acceptable as the face earlier obtained shouldn't repeat an so on...)

Probability = 720/ (6^6) = 5/324

Hence, B is the correct answer. Regards!
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