What is the probability that the sum of two dice will yield a 7, and

What is the probability that the sum of two dice will yield a 7, and then when both are thrown again, their sum will again yield a 7? assume that each die has 6 sides with faces numbered 1 to 6.

A. 1/144
B. 1/36
C. 1/12
D. 1/6
E. 1/3

IMO B is the correct answer
Probability to get sum as 7=(6/36)=1/6
Now we can these two incidents as two independent events
so again probability of getting sum as 7=(6/36)=1/6

So total Prob =1/6 * 1/6=
1/36

In order to get the sum of the two dices to be 7, there are six combinations that satisfy this requirement.
1+6=7
2+5=7
3+4=7
4+3=7
5+2=7
6+1=7

Each combination has 1/36 of probability of occurrence, or in total 6/36 for the six combinations, which is 1/6.
In order to get the probability of this occurring twice in a row we multiply 1/6 *1/6 and we get 1/36.

The answer is B.

What is the probability that the sum of two dice will yield a 7, and then when both are thrown again, their sum will again yield a 7? assume that each die has 6 sides with faces numbered 1 to 6.

A. 1/144
B. 1/36
C. 1/12
D. 1/6
E. 1/3

Solution -
Rolling dices is an independent event.
The combinations to get 7 are (1,6), (6,1), (2,5), (5,2), (3,4), (4,3) and total combinations of both dices is 36.

The probability of getting 7 in first attempt is 6/36=1/6.

Probability of getting 7 again in second attempt = (1/6)*(1/6) = 1/36. ANS B

Total Outcomes for a throw = 6*6 = 36
Favarable outcomes for 7 = 6 [Cases are {1, 6}{2, 5}{3, 4}{4, 3}{5, 2}{6, 1}]

Probability of Getting 7 in first throw = 6/36 = 1/6

Probability of Getting 7 in Two consecutive throws = (1/6)*(1/6) = 1/36

Answer: Option B

The pairs of values that will provide us a total 7 are:

1,6
2,5
3,4
4,3
5,2
6,1

There are 6*6 = 36 pairs possible on a throw of 2 dice.

Thus probability of getting a 7 first time = 6/36 = 1/6

And, probability of getting a 7 second time = 1/6*1/6 = 1/36, B is thus the answer.
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i understand that on first attempt probablity is 1/6 .
but for second atttempt why are we mulitplying it and why not adding it.. ?
And means multiplication and OR means addition - Is it a thumb rule ??

Yes, 'AND' in probability means that the outcome has to satisfy both conditions at the same time. Multiplying both the probabilities denotes 'and'.
'Or' means that the outcome has to satisfy one condition, or the other condition. Addition of the probabilities denotes 'or'.


Manhatten Prep article on the same:

https://www.manhattanprep.com/gmat/blog ... ld-series/
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Possible Options : (1,6),(6,1),(2,5),(5,2),(3,4),(4,3) :so probability of sum =7 is 6/36 -->1/6

If the dices are rolled again ,P2=1/6

Total : P1XP2 =1/36

Option B

Regards,
Manish Khare

MANHATTAN GMAT OFFICIAL SOLUTION:

There are 36 ways in which 2 dice can be thrown (6x6 = 36). The combinations that yield a sum of 7 are 1 + 6, 2 + 5, 3 + 4, 4 + 3, 5 + 2, and 6 + 1: 6 different combinations. Therefore, the probability of rolling a 7 is 6/36, or 1/6. To find the probability that this will happen twice in a row, multiply 1/6 by 1/6 to get 1/36.

Answer: B.
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On the first roll, the probability of a rolling 7 is 1/6. Because of the total of 36 possible combinations, 6 pairings yield 7; 6/36 = 1/6
For this probability to occur again, just multiple the original probability by itself.
1/6 * 1/6 = 1/36

B

Answer: B

Total outcomes: 6*6=36
Probability of getting a 7 first time: 6/36 (dice results can be: (1,6), (2,5),(3,4),(4,3),(5,2),(6,1))
Second throw is independent of first throw. So the outcomes will be same as above.
Probability of getting a 7 second time: 6/36

Total probability: (6/36)*(6/36) = 1/36

We need to find What is the probability that the sum of two dice will yield a 7, and then when both are thrown again, their sum will again yield a 7? assume that each die has 6 sides with faces numbered 1 to 6.

As we are rolling two dice => Number of cases = \(6^2\) = 36

Now for the sum to be 7 we need to find out what comes in both the dice roll. Following outcomes will yield 7 as the sum
(1,6), (2,5), (3,4), (4,3), (5,2), (6,1) = 6 outcomes

=> Probability that the sum of two dice will yield a 7 = \(\frac{6}{36}\) = \(\frac{1}{6}\)

Now, if we repeat this and again we have the same condition then in second case also we will get probability as \(\frac{1}{6}\)

Probability that these events will happen one after the other = Product of their probabilities = \(\frac{1}{6}\) * \(\frac{1}{6}\) = \(\frac{1}{36}\)

So, Answer will be B
Hope it helps!

Watch the following video to learn How to Solve Dice Rolling Probability Problems


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If we roll 2 dices in 1 turn, and the 2 dices are identical then a {1;6} and a {6;1}, etc., will not make a difference. How can we calculate it in that case then?
Should I assume such scenario will not happen in the test, if they dont call it out explicitly in the instruction (as per in this question)?
Hope someone can help