Q:

Expected Value Word Problem -Roulette Wheel

The probability a roulette wheel stops on a red number is 18/37.
If it lands on red and you bet red, then you receive double you bet (including your bet).
If you bet $1 on 10 consecutive plays, what is the probability that you make a profit?
My answer:
To make a profit, you need 5 wins and 5 loses.
Let X be the number of wins, X ~ Binomial(10, 18/37).
$$\Bbb P(5) = \binom{10}{5}(18/37)^5(19/37)^{5}$$
I found this to be equal to .245 which I have been told is incorrect.
I can't find any flaw in my thinking, any clue would be appreciated!
Thank you!

A:

You want to find P(X > 5), not P(X = 5).

