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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.

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All your arithmetic is correct. The problem is a logical error in the opening statement:

"The probability of win at least 1 prize, playing 3 times is the same of no win, that is: $\left(\frac{2}{3}\right)^3=\frac{8}{3^3}$"

It is not the same. Winning at least once and not winning at all are compliments. The outcomes are mutually exclusive and encompass all possibilities. Therefore, their probabilities should sum to 1. Solving the resulting equation gives the shortcut you are looking for.

$$P(A) + P(A^{c}) = 1$$

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  • $\begingroup$ Yes, was my mistake, thanks ! $\endgroup$ – Mattiu Apr 23 at 22:55
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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 $1$.

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