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If $Z$ represents number of times $6$ appeared in two independent throws of a die, what would be the probability of $A =${maximum of two throws was $2$ and $Z = 0$}. $Z = 0$ means $6$ appeared $0$ time.

-My try-

Probability of $6$ appears $ 0$ time = $\frac{25}{36}$

maximum of two throws would be $(1,2)$ or $(2,1)$ or $(2,2)$.

So there are $3$ ways to get maximum of two throws was $2$.

Therefore $P(A) = \frac3{36} \times \frac{25}{36}$ ???

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The events aren't independent. You multiply the probability that the max was two by the probability that there were no sixes given that the max was two. The latter probability is $1$. – David Mitra Jan 25 '13 at 3:15
up vote 1 down vote accepted

I am not sure that you are reporting the problem correctly. The maximum of the two throws is $2$ and there are no $6$ precisely if the maximum of the two throws is $2$. So the probability is $\dfrac{3}{36}$.

Maybe you were asked for the probability that the maximum of the throws is $2$ given that there was no $6$. Then by using the ordinary conditional probability procedure, we find that the probability the maximum is $2$ given there is no $6$ is $\dfrac{\frac{3}{36}}{\frac{25}{36}}$, which is $\dfrac{3}{25}$.

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Thank you very much! – hibc Jan 25 '13 at 4:08
By the way if p(A) = (max of throw is 2 | X = 0) then, P(A) = ( max of throw is 2 and X = 0) / (X=0). Then P(A) = ((3/36)(25/36)) / (25/36) ?? – hibc Jan 25 '13 at 4:22
Sure, that simplifies to $\frac{3}{36}$, the answer we got from much simpler considerations. – André Nicolas Jan 25 '13 at 4:25

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