You play a game of chance that you can either win or lose (there are no other
possibilities) until you lose. Your probability of losing is p = 0.64. Therefore your probability of winning is 1-0.64 = 0.36
What is the probability that it
a. takes 3 games until you lose = .0829
Let W = "win" and L = "loss"
P(W & W & L) = (0.36)(0.36)(0.64) = (0.36)2(0.64) = 0.082944
b. takes 6 games until you lose = .0039
P(W&W&W&W&W&L) = P(W)∙P(W)∙P(W)∙P(W)∙P(W)∙P(L) =
(0.36)(0.36)(0.36)(0.36)(0.36)(0.64) = (0.36)5(0.64) = 0.0038698353
c. Find the mean = ?????
That's the sum of all the expected number of times you win before you lose. That could be either 0 or 1 or 2 or 3 or 4 or …
The expected average number of times you win is the sum of all possible numbers of wins times the probability of winning than number of times.
[Notice that you may win 1000000 times and don't lose until the
1000001st time you play!!! So there are an infinite number of possible ways to play.]
Expectation of 0 wins (loss the 1st time) = 0∙(0.64) = 0
Expectation of 1 win = 1∙(0.36)(0.64)
Expectation of 2 wins = 2∙(0.36)2(0.64)
Expectation of 3 wins = 3∙(0.36)3(0.64)
Expectation of 4 wins = 4∙(0.36)4(0.64)
. . .
. . .
That's an infinite series we must sum.
We use the formula
d. Find the standard deviation = ????
That's a toughie. It is similar but a whole lot harder unless there's a
simpler formula for standard deviation than the basic definition.
Edwin