# Probability game of independent events

In one game, probability that a gamer scores win is 1/4. If X is number of tries( games ) that gamer has done in order to achieve score:

(NOTE!: every try is independent from the other)

a) Define the probability that gamer scored a win in 3 tries.

b) Define the probability that gamer will have at least three failures before he scores 2 wins.

I have tried to do it this way, but I am not sure if it is OK. a) P(win in 3 tries) = 1/4*1/4*1/4 = 1/64. b) for this task I have no idea from where to start.( I need great help here )

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What is the difference between the third try being a win and the first three tries are all wins? Hint: the probability of the latter is the $\frac{1}{4}\times\frac{1}{4}\times\frac{1}{4}$ that you calculated. Are there any other possibilities for what happened on the first two tries while still keeping the won on the third try? Or is it that a win occurs on the third try if and only if the first three tries are wins? – Dilip Sarwate Jul 2 '12 at 19:25

HINTS:

(a) The games tries are independent, so the probability that he scores a win on the third try is the same as the probability that he scores a win on the first try. What is that probability? (What you calculated is the probability that he wins all of his first three tries.)

(b) Think of it as a race: which comes first, his second win or his third failure? You want the probability that the third failure comes before the second win, but you might find it easier to calculate the probability that he scores $2$ wins before he has three failures and subtract that from $1$. How can he score $2$ wins before his third loss? Using W for a win and L for a loss, some possibilities are WW, WLW, LWW, WLLW, LWLW, and LLWW.

1. Are there any others?
2. Calculate the probability of each of the possibilies. For instance, $\Bbb P(\text{WLW})=\frac14\cdot\frac34\cdot\frac14$.
3. What do you have to do with these probabilities to get the total probability of scoring $2$ wins before his third loss?
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I have misprinted it under a, it was supposed to be in 3 tries, not in 3th try. For B, thanks very much, very good logically explained! :) – Takarakaka Jul 2 '12 at 19:31
I have done under B, I have done addition of all possible outcome for 2 wins before 3 failures, and it is ~0.22 ( P(WW)+P(WLW)+P(LWW)+P(WLLW)+P(LWLW)+P(LLWW) ) – Takarakaka Jul 2 '12 at 19:36
Under A I have calculated that three situations occur where win can be achieved in P(WLL), P(LWL) and P(LLW). So I have done addition over all those three possibilites and gained result of 3/4! :) – Takarakaka Jul 2 '12 at 19:40