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Probability question with a die?
Suppose that a regular die has been altered so that the faces are 1, 2, 3, 4, 5, and 5. If the die is tossed five times, what is the probability that the numbers recorded are 1, 2, 3, 4, and 5 in ANY order?
1 Answer
- IanLv 79 years agoFavorite Answer
First find the probability of 1, 2, 3, 4, 5 in this order, and then multiply by the number of ways of permuting 5 different numbers.
P(1, 2, 3, 4, 5 in this order)
= P(1 on 1st toss)P(2 on 2nd toss)P(3 on 3rd toss)P(4 on 4th toss)P(5 on 5th toss)
= (1/6)(1/6)(1/6)(1/6)(2/6)
= 1/3888.
There are 5! = 120 ways of permuting 5 numbers.
So P(1, 2, 3, 4, 5 in any order) = (1/3888)(120) = 5/162.
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