Wu rolls two fair, six-sided dice. What is the probability

[swerve]

Wu rolls two fair, six-sided dice. What is the probability that Wu rolls at least one five but no sixes?

A. 5/36
B. 275/1296
C. 2/9
D. 1/4
E. 5/18

The OA is D.

Please, can any expert explain this PS question for me? I tried to solve it but I can't get the correct answer and I would like to know how to solve it in less than 2 minutes. I need your help. Thanks.

[regor60]

Rolling at least one five and no sixes means:

one die with a 5 and the other with 1,2,3,4 OR

both dice with a 5

Because both of the above meet the criteria, you add their probabilities

Probability of a 5 on one die is 1/6. Probability of 1,2,3,4 on the other is 4/6 = 2/3. Multiplied together is 2/18.

Now because there are 2 ways for this to happen, the probability is 2x2/18 = 4/18 = 8/36

The probability of rolling two 5's is 1/6x1/6 = 1/36.

Adding 8/36 +1/36 = 9/36 = [spoiler]1/4, D[/spoiler]

[[email protected]]

Hi swerve,

We're told that Wu rolls two fair, six-sided dice. We're asked for the probability that Wu rolls AT LEAST one five but NO sixes. This question is all about 'probability math'; you might find it helpful to break the prompt into two smaller calculations:

Based on the types of outcomes that we are trying to achieve, there are two calculations to consider:

(1, 2, 3 or 4 on the 1st die)(5 on the 2nd die) = (4/6)(1/6) = 4/36
(5 on the 1st die)(1, 2, 3, 4 or 5 on the 2nd die) = (1/6)(5/6) = 5/36

Total Probability = 4/36 + 5/36 = 9/36 = 1/4

Final Answer: D

GMAT assassins aren't born, they're made,
Rich

[Scott@TargetTestPrep]

Wu rolls two fair, six-sided dice. What is the probability that Wu rolls at least one five but no sixes?

A. 5/36
B. 275/1296
C. 2/9
D. 1/4
E. 5/18

We can can break this up into scenarios:

Scenario 1:

P(5 on the first die and 1,2,3, or 4 on the second die)

1/6 x 4/6 = 4/36 = 1/9

Scenario 2:

P(1,2,3, or 4 on the first die and 5 on the second die)

4/6 x 1/6 = 4/36 = 1/9

Scenario 3:

P(5 on both dice)

1/6 x 1/6 = 1/36

Thus, the final probability is:

1/9 + 1/9 + 1/36 = 4/36 + 4/36 + 1/36 = 9/36 = 1/4.

Alternate Solution:

We see that of the 6 x 6 = 36 possible outcomes on the roll of two dice, only the outcomes of (5,1), (5,2), (5,3), (5,4); (1,5), (2,5), (3,5), (4,5) and (5,5) satisfy the required condition; a total of 9 outcomes. Thus, the probability that he rolls at least one 5 but no 6 is 9/36 = 1/4.

Answer: D