Sorry, full question: If I roll 5 standard 6-sided dice and multiply the number on the face of each die, what is the probability that the result is a composite number?
Can anyone help with this probability problem? Thanks so much!! :D
Can anyone help with this probability problem? Thanks so much!! :D
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As has been said, p(product of rolls is composite) = 1 - p(product of rolls is prime). It is easier to work out the probability of getting a prime number from the product of the 5 rolls.
There are 6^5 = 7776 ways of rolling 5 dice.
The only way that the product of the 5 rolls will be prime is for all but one of the dice to show a 1 and for the other die to show a prime (i.e. 2, 3 or 5). Otherwise, if two of the rolls are not equal to 1, the product of the rolls will be the product of two other numbers, i.e. a composite number. Hence there are 15 ways of rolling 5 dice so the product is prime, which are:
1,1,1,1,2
1,1,1,2,1
1,1,2,1,1
1,2,1,1,1
2,1,1,1,1
1,1,1,1,3
1,1,1,3,1
1,1,3,1,1
1,3,1,1,1
3,1,1,1,1
1,1,1,1,5
1,1,1,5,1
1,1,5,1,1
1,5,1,1,1
5,1,1,1,1
[Again, note that 1,1,1,1,1 doesn't count since their product is 1 which is not considered prime.]
Hence, the probability that the product of the rolls is prime is 15/7776.
Hence, p(composite) = 1- p(prime) = 1 - 15/7776 = 7761/7776 or about 0.998071 (99.8%.)
There are 6^5 = 7776 ways of rolling 5 dice.
The only way that the product of the 5 rolls will be prime is for all but one of the dice to show a 1 and for the other die to show a prime (i.e. 2, 3 or 5). Otherwise, if two of the rolls are not equal to 1, the product of the rolls will be the product of two other numbers, i.e. a composite number. Hence there are 15 ways of rolling 5 dice so the product is prime, which are:
1,1,1,1,2
1,1,1,2,1
1,1,2,1,1
1,2,1,1,1
2,1,1,1,1
1,1,1,1,3
1,1,1,3,1
1,1,3,1,1
1,3,1,1,1
3,1,1,1,1
1,1,1,1,5
1,1,1,5,1
1,1,5,1,1
1,5,1,1,1
5,1,1,1,1
[Again, note that 1,1,1,1,1 doesn't count since their product is 1 which is not considered prime.]
Hence, the probability that the product of the rolls is prime is 15/7776.
Hence, p(composite) = 1- p(prime) = 1 - 15/7776 = 7761/7776 or about 0.998071 (99.8%.)
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there are 3 primes on a 6 sided die, 2, 3, and 5. 1 is a factor of everything.
so if 4 of the dice come up 1,2,3,5, then the 5th die must either repeat one of those 4, which would give you a composite (except for the 1), or show a composite, which also gives you a composite product.
so the only way NOT to get a composite product is to roll 1-1-2-3-5
there are 6!/2! 360 ways to arrange that, out of 6^6 = 46656 possible outcomes,
so probability of prime outcome is 0.007716,
probability of composite outcome is 1 – 0.007716 = 0.992284
so if 4 of the dice come up 1,2,3,5, then the 5th die must either repeat one of those 4, which would give you a composite (except for the 1), or show a composite, which also gives you a composite product.
so the only way NOT to get a composite product is to roll 1-1-2-3-5
there are 6!/2! 360 ways to arrange that, out of 6^6 = 46656 possible outcomes,
so probability of prime outcome is 0.007716,
probability of composite outcome is 1 – 0.007716 = 0.992284
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I'll make a stab at this. The only way it would not be a composite number is if you roll all ones or
roll a prime number once and the rest all ones. 2, 3, and 5 are prime numbers. to get all ones,
each one having a one-sixth chance of appearing, is (1/6)^5. To roll a 2 and 4 ones is (1/6)^5 also, as are the options with 3 and 5. Add these four and get 4 (1/6)^5.Subtract this from 1 and I get
.999485597. I am not sure this is right, but that is a start.
roll a prime number once and the rest all ones. 2, 3, and 5 are prime numbers. to get all ones,
each one having a one-sixth chance of appearing, is (1/6)^5. To roll a 2 and 4 ones is (1/6)^5 also, as are the options with 3 and 5. Add these four and get 4 (1/6)^5.Subtract this from 1 and I get
.999485597. I am not sure this is right, but that is a start.