2018 AMC 10 B

Complete problem set with solutions and individual problem pages

Problem 9 Easy

The faces of each of 7 standard dice are labeled with the integers from 1 to 6. Let P be the probability that when all 7 dice are rolled, the sum of the numbers on the top faces is 10. What other sum occurs with the same probability P? (2018 AMC 10B Problem, Question#9)

  • A.

    13

  • B.

    26

  • C.

    32

  • D.

    39

  • E.

    42

Answer:D

It can be seen that the probability of rolling the smallest number possible is the same as the probability of rolling the largest number possible, the probability of rolling the second smallest number possible is the same as the probability of rolling the second largest number possible, and so on. This is because the number of ways to add a certain number of ones to an assortment of 7 ones is the same as the number of ways to take away a certain number of ones from an assortment of 7~6s. So, we can match up the values to find the sum with the same probability as 10. We can start by noticing that 7 is the smallest possible roll and 42 is the largest possible role. The pairs with the same probability are as follows: (7, 42), (8, 41), (9, 40), (10, 39), (11, 38)\cdots

However, we need to find the number that matches up with 10. So, we can stop at (10, 39) and deduce that the sum with equal probability as 10 is 39. So, the correct answer is \rm (D)~39, and we are done.

Let's call the unknown value x . By symmetry, we realize that the difference between 10 and the minimum value of the rolls is equal to the difference between the maximum and x. So, 10-7=42-x, x=39 and our answer is \rm (D)~39.

For the sums to have equal probability, the average sum of both sets of 7 dies has to be (6+1) \times 7= 49. Since having 10 is similar to not having 10, you just subtract 10 from the expected total sum. 49 -10 = 39 so the answer is  \rm (D)~39.

The expected value of the sums of the die rolls is 3.5*7=24.5, and since the probabilities should be distributed symmetrically on both sides of 24.5, the answer is 24.5+24.5-10=39, which is \rm (D)~39.

Calculating the probability of getting a sum of 10 is also easy. There are 3 cases:

Case 1: \left\{1,1,1,1,1,1,4\right\},

\frac{7!}{6!}=7 cases,

Case 2: \left\{1,1,1,1,1,2,3\right\},

\frac{7!}{5!}=6*7=42 cases,

Case 3: \left\{1,1,1,1,2,2,2\right\},

\frac{7!}{4!3!}=5*7=35 cases,

The probability is \frac{84}{6^7}=\frac{14}{6^6}.

Calculating 6^6:

6^6=(6^3)^2=216^2=46656,

Therefore, the probability is \frac{14}{46656}=\frac{7}{23328}.

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