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So the problem is: What is the probability of rolling eleven dice such that their product is prime. The dice is numbered from 1 to 6 and there is an equal chance of getting each number.

So in order for the product to be prime, all but one of the numbers must be 1. The number that is not 1 must be either 2, 3, or 5. So the total number of ways that 2, 3, or 5 can be formed through multiplication is 3*11 = 33. I then divided 33 by the total number of products that can be formed, or $6^{11}$.

My original solution yields basically the same numerical value, but I'm doubtful of it's correctness. By the binomial probability theorem, the probability of rolling a 1 ten times and a number besides 1 is

$$_{11}C_{10} \left(\frac{1}{6}\right)^{10}\left(\frac{5}{6}\right)^{1} $$

Given that the remaining number is not 1, the probability of rolling 2, 3, or 5 is $\frac{3}{5}$. Hence we multiply them together to get the probability $$_{11}C_{10} \left(\frac{1}{6}\right)^{10}\left(\frac{5}{6}\right)^{1}\left(\frac{3}{5}\right)$$

This problem was originally a test question, but many of my peers got a different answer of $$ \left(\frac{(3*11)}{6^{11}}\right)\left(\frac{11!}{10!}\right)$$

If their answer is correct could someone please explain to me why mines is incorrect? I would also greatly appreciate it if you could explain how to get the correct answer.

pjs36
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Gates VVang
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1 Answers1

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As the dice are fair, it is easier to confirm your answer by approaching the problem via counting principles.

How many sequences of length 11 with entries being natural numbers 1 through 6 are there? This will be our sample space (all base6 sequences of length 11). By multiplication principle, we have that there are $6^{11}$ such sequences.

How many of these sequences has the product of entries in the sequence as a prime number. As you correctly noted, this only happens if all but one of the entries is a 1 and the not-one entry is prime (2,3 or 5). Set up a multiplication principle argument as follows:

  • Pick the location of the not-one number (11 choices)
  • Pick the prime used for the not-one number (3 choices)

All other spaces then will be ones. As such, there are 33 possible such sequences.

By definition of probability in an equiprobable sample space, the answer is then $\frac{3\cdot 11}{6^{11}}$

Your answer is equivalent: $~_{11}C_{10} \left(\frac{1}{6}\right)^{10}\left(\frac{5}{6}\right)^{1}\left(\frac{3}{5}\right)=11\cdot (\frac{1}{6^{10}})\cdot \frac{5}{6}\cdot \frac{3}{5} = \frac{11\cdot 3}{6^{11}}$

Their answer on the other hand, overcounts by a factor of 11. I cannot explain where their answer comes from, but perhaps they got confused and multiplied by 11 for the location of the not-one twice (once in the fraction on the left and once in the fraction on the right).

JMoravitz
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