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Hi
I need help with the following question.
Four people each will roll a fair die once. What is the probability that at least two will roll the same number?
A) 50%
B) 65%
C) 80%
D) 72%
E) 40%
The problem can easily be solved by making all the people roll the same number then subtracting the result from 1. However, I wanted to understand the problem by adding up all the cases i.e two roll the same number or three roll the same number etc. but i am not getting the correct answer. Can any expert help me find the mistake?
Case 1 Two people throw the same number and the other two each roll a different number. e.g 4456
6/6 * 1/6 *5/6 * 4/6 *4!/3!2! I am not sure how to calculate all the permutations of this. To arrange four numbers i multiplied by 4! then divided by 3! to eliminate double counting and again by 2! to remove the same digit cases i.e 44 55.
Case 2 two roll the same number and the other two roll the same but different from the one that was rolled by the first two people. e.g 3344
6/6* 1/6* 5/6* 1/6 *4/2! 2! 2! Here I arrange the four numbers then divided by 2! three times to remove double digits already arranged in the product and to eliminate two same digit groups. again i am not sure if this is correct way to count all permutations of this case.
Case 3 three roll the same number and the third rolls a different number eg 4445
6/6 * 1/6*1/6*5/6 * 4!/ 2!*3!
Case 4 All roll the same number
this is easy 6/6^4
Adding up all the cases does not yield the right answer. Please help me find the mistake. Thanks
Jac
I need help with the following question.
Four people each will roll a fair die once. What is the probability that at least two will roll the same number?
A) 50%
B) 65%
C) 80%
D) 72%
E) 40%
The problem can easily be solved by making all the people roll the same number then subtracting the result from 1. However, I wanted to understand the problem by adding up all the cases i.e two roll the same number or three roll the same number etc. but i am not getting the correct answer. Can any expert help me find the mistake?
Case 1 Two people throw the same number and the other two each roll a different number. e.g 4456
6/6 * 1/6 *5/6 * 4/6 *4!/3!2! I am not sure how to calculate all the permutations of this. To arrange four numbers i multiplied by 4! then divided by 3! to eliminate double counting and again by 2! to remove the same digit cases i.e 44 55.
Case 2 two roll the same number and the other two roll the same but different from the one that was rolled by the first two people. e.g 3344
6/6* 1/6* 5/6* 1/6 *4/2! 2! 2! Here I arrange the four numbers then divided by 2! three times to remove double digits already arranged in the product and to eliminate two same digit groups. again i am not sure if this is correct way to count all permutations of this case.
Case 3 three roll the same number and the third rolls a different number eg 4445
6/6 * 1/6*1/6*5/6 * 4!/ 2!*3!
Case 4 All roll the same number
this is easy 6/6^4
Adding up all the cases does not yield the right answer. Please help me find the mistake. Thanks
Jac
