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A gambler rolls three fair six-sided dice. What is the

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A gambler rolls three fair six-sided dice. What is the

by VJesus12 » Sat Aug 25, 2018 2:24 am

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A gambler rolls three fair six-sided dice. What is the probability that two of the dice show the same number, but the third shows a different number?

A) 1/8
B) 5/18
C) 1/3
D) 5/12
E) 5/6

The OA is the option D.

Is there a fast way to solve it? Or a strategic way? I would appreciate any help. Please.

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by Brent@GMATPrepNow » Sat Aug 25, 2018 5:11 am
VJesus12 wrote:A gambler rolls three fair six-sided dice. What is the probability that two of the dice show the same number, but the third shows a different number?

A) 1/8
B) 5/18
C) 1/3
D) 5/12
E) 5/6
Let's first calculate P(same, same, different)
P(same, same, different) = P(1st roll is ANY value AND 2nd roll matches 1st roll AND 3rd roll is different from first 2 rolls)
= P(1st roll is ANY value) x P(2nd roll matches 1st roll) x P(3rd roll is different from first 2 rolls)
= 6/6 x 1/6 x 5/6
= 5/36
So, P(same, same, different) = 5/36

However, this is not the only way to get 2 sames and 1 different.
There's also: same, different, same as well as different, same, same

Applying similar logic, we know that P(same, different, same) = 5/36
And P(different, same, same) = 5/36

So, P(2 same rolls and 1 different) = P(same, same, different or different, same, same or same, different, same)
= P(same, same, different ) + P(different, same, same) + P(same, different, same)
= 5/36 + 5/36 + 5/36
= 15/36
= 5/12

Answer: D

Cheers,
Bret
Brent Hanneson - Creator of GMATPrepNow.com
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by swerve » Sat Aug 25, 2018 9:37 am
Hi VJesus12!

The total number of possible way: 6*6*6=216.
3 discs will appear in any one of the following arrangements: AAA, AAB, ABC, where AAA=All the same, AAB=Two are the same, ABC=All three different.

Now, the total number of AAA = 6.
The total number of ABC = 6*5*4 = 120.
Therefore, the total number of AAA, ABC = 126.

So, the total number of AAB = 216 - 126 = 90.
Probability of AAB = 90/216 = 5/12.

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by Jeff@TargetTestPrep » Tue Sep 04, 2018 3:30 am
VJesus12 wrote:A gambler rolls three fair six-sided dice. What is the probability that two of the dice show the same number, but the third shows a different number?

A) 1/8
B) 5/18
C) 1/3
D) 5/12
E) 5/6
The total number of outcomes when 3 dice are rolled is 6 x 6 x 6 = 216.

We can use the following formula:

P(two of the dice show the same number, but the third shows a different number) =

1 - P(all three dice show the same number) - P(all three dice show a different number)

Now,

P(all three dice show a different number) = 6/6 x 5/6 x 4/6 = 20/36

and

P(all three dice show the same number) = 6/6 x 1/6 x 1/6 = 1/36

Thus,

P(two of the dice show the same number, but the third shows a different number) = 1 - 20/36 - 1/36 = 1 - 21/36 = 15/36 = 5/12

Answer: D

Jeffrey Miller
Head of GMAT Instruction
[email protected]

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