Online Course Discussion Forum

AMC12B FALL 2021 #11

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AMC12B FALL 2021 #11
by Jialin Wang - Sunday, October 30, 2022, 4:48 PM

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 $4?$

I am really confused how things work here.

 I tried to do complementary counting. There are a total of 6^6 ways, which is the denominator. And for the numerator, one of the dice can be 1,2,3,5,or 6,  and the rest have to be odd, so it's 5*3^5 ways. Then I tried to do (6^6-5*3^5)/6^6, which is 187/192.

 It was not the correct answer, which is  59/64. 

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Re: AMC12B FALL 2021 #11
by Dr. Kevin Wang - Wednesday, November 2, 2022, 3:00 PM

The idea of complementary counting is correct.

To count the cases that the product is NOT divisible by $4$, there can be at most one factor of $2$.  So far so good.

However, it is not as simple as you stated.  One of the dice can be 1,2,3,5, or 6---this is true, but you also need to choose that die, so you need to multiply by $6$.  That is not all the problems here.  When this die has number 1, 3, or 5, it is odd, same as the other dice, so there is a lot of overcounting here.

To count it in a clean way, consider the cases (1) All dice show odd numbers, and (2) One of the dice is 2 or 6, and all others are odd.  Then you can subtract both cases from the total.

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Re: AMC12B FALL 2021 #11
by Jialin Wang - Tuesday, November 1, 2022, 8:17 PM

Thank you!