Stats/ Probability
posted by denim .
Repeated die toss experiment. A fair die is tossed repeatedly, and the sequence of numbers that turn up is recorded, until a 1 is
obtained. When a 1 is obtained, the experiment is over, but the 1 is included in the sequence. Thus every outcome is a sequence of integers ending
in 1. Let X be the sum of the sequence of numbers obtained in the experiment. So the smallest possible value of X is 1, corresponding to the outcome
in which a 1 is obtained on the first toss.
Compute P(X=5).

Probability of NOT getting a 1 = 5/6, getting a 1 = 1/6.
If the events are independent, the probability of both/all events occurring is determined by multiplying the probabilities of the individual events.
For five tosses:
(5/6)^4 * 1/6 = ?
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