2021 AMC 10 B Fall

Complete problem set with solutions and individual problem pages

Problem 14 Medium

Una rolls 6 standard 6 -sided dice simultaneously and calculates the product of the 6 numbers obtained. What is the probability that the product is divisible by 4 ?(2021 AMC Fall 10B, Question #14)

  • A.

    \frac{3}{4}

  • B.

    \frac{57}{64}

  • C.

    \frac{59}{64}

  • D.

    \frac{187}{192}

  • E.

    \frac{63}{64}

Answer:C
Solution

We will use complementary counting to find the probability that the product is not divisible by 4. Then, we can find the probability that we want by subtracting this from 1 . We split this into 2 cases.

Case 1: The product is not divisible by 2 . We need every number to be odd, and since the chance we roll an odd number is \frac{1}{2}, our probability is \left(\frac{1}{2}\right)^{6}=\frac{1}{64}.

Case 2 : The product is divisible by 2 , but not by 4 . We need 5 numbers to be odd, and one to be divisible by 2 , but not by 4 . There is a \frac{1}{2} chance that an odd number is rolled, a \frac{1}{3} chance that we roll a number satisfying the second condition (only 2 and 6 work), and 6 ways to choose the order in which the even number appears. Our probability is \left(\frac{1}{2}\right)^{5}\left(\frac{1}{3}\right) \cdot 6=\frac{1}{16}. Therefore, the probability the product is not divisible by 4 is \frac{1}{64}+\frac{1}{16}=\frac{5}{64}. Our answer is 1-\frac{5}{64}= (C) \frac{59}{64}

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