There are 6 dice. The probability of each number is the same with an exception. The number 4 has a 2/7 chance of occurring while everything else has a 1/7
One of my main issues is that I'm having trouble distinguishing if I should be using Permutations/Arrangements or Combinations
What is the probability that all six dice show the same number?
The chance you get all six dice to show the same result for any number other than 4 is (1/7)^6 so we count all of them so we do 5 * (1/7)^6 which is 5/117649. We also need to factor in the chance that 4 is the number all the die land on which has a chance of (2/7)^6. The answer is the addition of these probabilities which is 69/117649.
Is this correct?
What is the probability that four of the dice show the number five and the other two show the number four?
(6 choose 4) comes from the fact that there are C(6,4) ways that four of the dice can get the number 5 and the probability of that happening is (1/7) ^4. The probability that the other two dice leftover is (2/7)^2. My final result ends up being 60/117649 Is this correct?
What is the probability that no two dice show the same number, i.e., that you roll the numbers 1, 2, 3, 4, 5, and 6?
(6 choose 5) * (1/7)^5 * 2/7 = 2/117649
What is the probability that exactly three of the dice show the number five?
I did a binomial distribution giving me 6 choose 3 * (1/7)^3 * (6/7)^3 =4320/117649 = .0018