The probability of Sam beating Abby in a game of cards is 0.36. Abby and Sam decide to play a game everday for n days. What is the fewest number of games they need to play to ensure the probability of Sam winning at least once is greater than 0.85.
Is the answer 5?
I got 5 too.
This is how I did it.
Pr(success; eg Sam beating Abby) = 0.36
Pr(X greater than or equal to 1) > 0.85
1 - Pr(X less than 1) > 0.85
- Pr (X less than 1) > -.015
Pr (X less than 1) < 0.15 (flipping because of "negative")
Another way of writing Pr (X less than 1) is Pr (X=0).
Pr (X=0) < 0.15
using binomial theorem, we get nC0 * (0.36)^0 * (0.64)^n <0.15
(0.64)^n < 0.15
However, you calculator cannot help the above. Type (0.64)^n = 0.15, to get n = 4.2509
However, you can't have 4.2509 days since they are playing once a day. You can't round down, because this would not ensure the probability to be greater than 0.85. Therefore you round up, getting n = 5
