Four people each roll a fair die once.
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17 Jul 2022, 03:32
Please note that there 4 possible different cases:
Case 1: Anyone will theow the same number (e.g., 6543)
Case 2: Two people will throw the same number (e.g., 6654)
Case 3: Three people will throw the same number (e.g., 6665)
Case 4: Everyone will throw the same number (e.g, 6666)
Since P(at least two people will roll the same number) is equal to the sum of the probabilities of case 2, case 3 and case 4, the easiest way to solve this problem is to calculate the probability of case 1 and get the difference to 1. In this way, we calculate only one probability instead of three.
Thus, P(at least two people will roll the same number) = 1 - P(anyone will throw the same number)
P(anyone will throw the same number) = 5/6*4/6*3/6 = 0,27777...
You can do it this way because it doesn't matter which will be the value of the first throw. Another way to do it would be:
Case 1st throw = 1: 1/6*5/6*4/6*3/6
Case 1st throw = 2: 1/6*5/6*4/6*3/6
Case 1st throw = 3: 1/6*5/6*4/6*3/6
Case 1st throw = 4: 1/6*5/6*4/6*3/6
Case 1st throw = 5: 1/6*5/6*4/6*3/6
Case 1st throw = 6: 1/6*5/6*4/6*3/6
If you sum these 6 cases you will get 6/6*5/6*4/6*3/6, which is equal to 5/6*4/6*3/6 = 0,27777...
Finally, if we do the subtraction 1 - 0,2777 we get 0,72
Answer A since 0,72 > 0,70