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When 2 dices tossed, what is the probability that 2 numbers

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by Max@Math Revolution » Sun Feb 14, 2016 10:41 pm
When 2 dices tossed, what is the probability that 2 numbers faced have difference 2?

A. 1/9
B. 2/9
C. 1/3
D. 4/9
E. 5/9


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by Brent@GMATPrepNow » Mon Feb 15, 2016 7:14 am
Max@Math Revolution wrote:When 2 dices tossed, what is the probability that 2 numbers faced have difference 2?

A. 1/9
B. 2/9
C. 1/3
D. 4/9
E. 5/9
P(dice have difference of 2) = P(1st die is 1 and 2nd die is 3 OR 1st die is 2 and 2nd die is 4 OR 1st die is 3 and 2nd die is 1 or 5 OR 1st die is 4 and 2nd die is 2 or 6 OR 1st die is 5 and 2nd die is 3 OR 1st die is 6 and 2nd die is 4 OR )
= P(dice have difference of 2) = P(1st die is 1 and 2nd die is 3) + P(1st die is 2 and 2nd die is 4) + P(1st die is 3 and 2nd die is 1 or 5) + P(1st die is 4 and 2nd die is 2 or 6) + P(1st die is 5 and 2nd die is 3) + P(1st die is 6 and 2nd die is 4)
= (1/6)(1/6) + (1/6)(1/6) + (1/6)(2/6) + (1/6)(2/6) + (1/6)(1/6) + (1/6)(1/6)
= 1/36 + 1/36 + 2/36 + 2/36 + 1/36 + 1/36
= 8/36
= 2/9
= B

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by Max@Math Revolution » Tue Feb 16, 2016 5:34 pm
When 2 dices tossed, what is the probability that 2 numbers faced have difference 2?

A. 1/9
B. 2/9
C. 1/3
D. 4/9
E. 5/9


--> The probability that 2 numbers facing have difference of 2 is (1,3),(2,4),(3,5),(4,6),(3,1),(4,2),(5,3),(6,4), which makes 8 cases. Then, 8C1/6C1*6C1=8/6*6=2/9. Therefore, the answer is B.
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