Question

Michael buys a ticket in the Tri-State Pick 3 lottery every day, always betting on 812. He will win something if the winning number contains 8, 1, and 2 in any order. Each day, Michael has probability 0.006 of winning, and he wins (or not) independently of other days because a new drawing is held each day. What is the probability that Michael’s first winning ticket comes on the 10th day?

Answer 1

0.00568 ANS

Explanation:

Since Michael has a .006 chance of winning on any given day,

Than his tries of not winning on any day are as:

1-.006=.994.

For Michael first winning ticket would be on 10 day, Michael should not win on the first 9 days, so then he has to win on the 10 day. For Michael not to succeed on the first 9 days, the probability is:

.994^9=0.94727801832

Hence for him to succeed on the 10 day, the probability is .006.

Now the probability of his first winning ticket being on the 10 day is:

.006*.994^9.=0.0056836681.