# Probability to win at least 1 time, playing 3 times.

I have this problem:

In a game, the probability of win is $$1/3$$ and of lose is $$2/3$$, ¿What is the probabilty of win at least 1 prize playing 3 times?

The probability of win at least 1 prize, playing 3 times is the same of no win, that is: $$(2/3)^3 = 8/3^3$$

Now, the long way would be:

The probability of winning at least 1 price is equal to:

P(win 1 prize) + P(win 2 prize) + P(win 3 prize), that is:

P(win 1 prize) have 3 different ways: WLL, LLW, LWL(W = win, L = lose) and the probabilty for each one is $$1/3 * 2/3 * 2/3 = (4/3^3)$$ and since they are 3 different ways, so is: $$3 *(4/3^3) = 4/3^2$$

P(win 2 prize) have 3 different ways: WWL, WLW, LWW. = $$3*(1/3 * 1/3 * 2/3) = 2/3^2$$

P(win 3 prize) have only one way: WWW = $$(1/3^3)$$

adding the probabilities: $$2/3^2 + 4/3^2 + 1/3^3 = 19/3^3$$, but this result is different of $$8/3^3$$ that is the result of no win.

Where is my mistake? Thanks in advance.

"The probability of win at least 1 prize, playing 3 times is the same of no win, that is: $$\left(\frac{2}{3}\right)^3=\frac{8}{3^3}$$"
$$P(A) + P(A^{c}) = 1$$
Probability of win at least one time isn't the same as probability of no win, it's one minus probability of no win. And indeed $$\frac{19}{3^3}$$ and $$\frac{8}{3^3}$$ sums to $$1$$.