GeminiHeat 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, sameApplying 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 - founder of Greenlight Test Prep