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A

I’ve got a long solution for this question, and think the answer is 5/54.

Answer A) Since, Favourable outcomes are

R B B No

3 1 1 1

4 1,2 2,1 3 (Always we could add the previous total)

5 1,2,3 3,2,1 6

6 1,2,3,4 4,3,2,1 10

Total fav outcomes is 20

Probability = 20/6*6*6

= 5/54

A for sure, I wish I could attach my scratch paper to show the whole process

C

Hello Chris

According to my calculations answer is A – 5/54

But it was time taking as I did it by listing technique. Kindly update us with right answer and time efficient solution.

two blues: 1 and 1 vs one red: 3,4,5,6 = Probability:1/6*1/6*4/6=4/6^3

two blues: 1 and 2 vs one red: 4,5,6 = Probability:1/6*1/6*3/6*2=6/6^3

two blues: 2 and 2 vs one red: 5,6 = Probability:1/6*1/6*2/6=2/6^3

two blues: 1 and 3 vs one red: 5,6 = Probability:1/6*1/6*2/6*2=4/6^3

two blues: 1 and 4 vs one red: 6 = Probability:1/6*1/6*1/6*2=2/6^3

two blues: 2 and 3 vs one red: 6 = Probability:1/6*1/6*1/6*2=2/6^3

We multiply by 2 when blues result in different values so they can be interchanged

As we calculate probability for OR we have to sum six fractions which brings 20/6^3 or 5/54. A

Assuming outcomes like (1,2) – (2,1) on blue dice to be same.

Possibilities on blue dice (1,1) , (1,2) , (1,3) , (2,2) and, (2,3) ; total 5 cases.

Corresponding possibilities on red dice = {3,4,5,6}, {4,5,6}, {5,6} and {6} so total of 4+3+2+1 = 10 cases.

Total number of cases, on blue dice = (6C1x6C1)/2! , 2! for symmetrical cases like (1,2) – (2,1) , (2,3) – (3,2) and so on. on red dice = 6C1

So, required probability = 10/( (6C1x6C1)/2!) x (6C1) ) = 5/54.

Therefore, A is the answer.

A… 5/54

A

D. As there are 2 blue dice, 6*6 gives the total number of possible outcomes = 36.

Favorable outcomes, the Red dice can have a maximum value of 6. Rolling two blue dice and having there sum < 6, below are the possible values.

1,1

1,2

1,3

1,4

2,1

2,2

2,3

So there are 7 favorable outcomes.

The answer should be D – 7/36

Please let me know if this is incorrect.

Your suggestions are welcome.

Wish you happy and a prosperous new year 🙂

Thanks,

Swaroop

You forgot about the red die: A conjuror will roll three dice. Though, mentioning of the right hand only for a red die does confuse (sorry Chris)))

And more, when the values of blue dice are not equal (e.g., 2 and 3 as opposed to 1 and 1) you are supposed to double that probability.

Accually, when you mentiones the possible outcomes, you forgot some:

1,1

1,2

2,1

1,3

3,1

1,4

4,1

2,2

2,3

3,2

That gives us a total of 10 favorable outcomes. Therefore, the answer should be 10/36 or 5/18.

what about the probability of red one

red blue1 blue2

3 1 1 1/6*1/6*1/6

4 1 1 1/6*1/6*1/6

4 2 1

4 1 2

.

.

. ———————-

the sum is 10/108=5/54

the answer is A

You are forgetting 3 cases of 3,2 3,1 & 4,1. Also 10 cases when red dice is showing 3, 4 or 5 and is still larger than sum of 2 blue. For example, red is 3 and blue are both 1. Correct answer is A.