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Practice problems on probability, permutation, combination

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by georgeanand » Thu Aug 07, 2014 1:31 am
Problem 1

A dice is rolled 3 times and sum of three numbers appearing on the uppermost face is15.
The chance that the first roll was a four is...
(1)2/5 (2)1/5 (3)1/6 (4) 2/7


Option 2 is the answer.

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by GMATGuruNY » Thu Aug 07, 2014 3:14 am
georgeanand wrote:Problem 1

A dice is rolled 3 times and sum of three numbers appearing on the uppermost face is15.
The chance that the first roll was a four is...
(1)2/5 (2)1/5 (3)1/6 (4) 2/7
List all of the ways that the 3 rolls could yield a sum of 15:
3+6+6
4+5+6
4+6+5
5+4+6
5+5+5
5+6+4
6+3+6
6+4+5
6+5+4
6+6+3
Of the 10 ways, only the 2 options in red satisfy the constraint that the first roll is 4.
Thus:
P(1st roll is 4) = 2/10 = 1/5.

The correct answer is B.
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by GMATinsight » Thu Aug 07, 2014 6:42 am
georgeanand wrote:Problem 1

A dice is rolled 3 times and sum of three numbers appearing on the uppermost face is 15.
The chance that the first roll was a four is...
(1)2/5 (2)1/5 (3)1/6 (4) 2/7
Total cases of 15 are as follows
{6, 6, 3}
{6, 5, 4}
{6, 4, 5}
{6, 3, 6}
{5, 6, 4}
{5, 5, 5}
{5, 4, 6}
{4, 6, 5} First Favorable case
{4, 5, 6} Second Favorable case
{3, 6, 6}

Total Cases = 10
Total Favorable cases = 2

Probability = 2/10 = 1/5

Answer: Option B
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