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I have a question regarding the order of the dice combinations

If the first dice rolls a 6 and the second dice rolls a 6, that is 1 combination greater than 11. However, doesn't that mean that we can reverse the order and get the same result?

Hence: 1/6 *1/6 *2 since they can be flipped. Why can we not do that and is there any material I can read to get a better grasp on this concept?
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Solution


Given:
    • Two dice are thrown simultaneously

To find:
    • The probability that the sum of the two numbers that turn up < 11

Approach and Working:
    • The probability that the sum of the two numbers that turn up < 11 = 1 – (the probability that the sum = 11 + the probability that the sum = 12)
    • Therefore, the probability that the sum of the two numbers that turn up < 11 =\(1 – (\frac{2}{6*6} + \frac{1}{6*6}) = 1 - \frac{3}{36} = 1 - \frac{1}{12} = \frac{11}{12}\)

Hence, the correct answer is Option B

Answer: B


The attached table comes in handy for quickly resolving problems involving sums of 2 and 3 dice. It is worth memorizing to save time. With these tables I can solve problems involving sums of 2 and 3 dices in less than 1 minute.
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When two dice are thrown simultaneously, what is the probability that the sum of the two numbers that turn up is less than 11?

A. 35/36
B. 11/12
C. 5/6
D. 1/6
E. 1/12

All possibilities except (5,6) ; (6,5) ; (6,6) where the sum of two numbers is > 11

Total possibilities, 6*6 =36, since with each number has 6 pairs

Thus P (A) = (36-3) /36 = 33/36 = 11/12 (Ans -B)
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Given that two dice are thrown simultaneously and we need to find what is the probability that the sum of the two numbers that turn up is less than 11?

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

Out of the 36 cases only three cases are there where the sum will be 11 or more. These cases are (5,6), (6,5) and (6,6)
=> For 36 - 3 = 33 cases we will have the sum less than 11

=> Probability that the sum will be less than 11 = \(\frac{33}{36}\) = \(\frac{11}{12}\)

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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