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# Twelve fair dice are rolled. What is the probability that the product

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Math Expert
Joined: 02 Sep 2009
Posts: 58454
Twelve fair dice are rolled. What is the probability that the product  [#permalink]

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24 Mar 2019, 23:16
1
3
00:00

Difficulty:

75% (hard)

Question Stats:

25% (02:15) correct 75% (02:17) wrong based on 16 sessions

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Twelve fair dice are rolled. What is the probability that the product of the numbers on the top faces is prime?

(A) $$(\frac{1}{12})^{12}$$

(B) $$(\frac{1}{6})^{12}$$

(C) $$2(\frac{1}{12})^{11}$$

(D) $$\frac{5}{2}(\frac{1}{6})^{11}$$

(E) $$(\frac{1}{6})^{10}$$

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Joined: 27 Sep 2018
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Re: Twelve fair dice are rolled. What is the probability that the product  [#permalink]

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24 Mar 2019, 23:48
Prime numbers 2,3,5
For a product to be prime one of the die should have prime number and all other should have 1

Prob ability of getting prime number =3/6=1/2
Prob ability of getting 1 =(1/6)^11
Number of ways these can occur =12C1=12

So , the product of the numbers on the top faces is prime= 1/2*(1/6)^11*12=(1/6)^10

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Re: Twelve fair dice are rolled. What is the probability that the product  [#permalink]

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24 Mar 2019, 23:57
Bunuel wrote:
Twelve fair dice are rolled. What is the probability that the product of the numbers on the top faces is prime?

(A) $$(\frac{1}{12})^{12}$$

(B) $$(\frac{1}{6})^{12}$$

(C) $$2(\frac{1}{12})^{11}$$

(D) $$\frac{5}{2}(\frac{1}{6})^{11}$$

(E) $$(\frac{1}{6})^{10}$$

so as to get product as prime 11 of 12 dice have to be 1 and the other can be either of 2,3,5
so
P = 2,3,5; 3/6 = 1/2
P 1= (1/6)^11
and 12 times its rolled so = 1/2 * (1/6)^11 * 12 = (1/6)^10
IMO E
Re: Twelve fair dice are rolled. What is the probability that the product   [#permalink] 24 Mar 2019, 23:57
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