We have two dice, one of which is loaded with faces numbered $\{1,1,2,3,4,5\}$. If we choose one of them and we launch it 5 times, let $X$ be the random variable "the number of times we get 1", calculates $P\{X>=2\}$ and $E[X]$? We have two cases, when we choose a normal die: $P\{X>=2\}=1-P\{X=0\}-P\{X=1\}=\frac{1526}{6^5}$, when we choose a loaded die: $P\{X>=2\}=1-P\{X=0\}-P\{X=1\}=\frac{131}{3^5}$, I have used the binomial random variable with parameters respectively of: $(n=5, p=\frac{1}{6})$ and $(n=5, p=\frac{2}{6})$. Now i think that i should join this two cases but i don't know how to do it and if it is correct.

Using the inclusion exlusion principle: $P\{X>=2\}=P\{X>=2\}_{N}+P\{X>=2\}_{L}-P(\{X>=2\}_{N}\{X>=2\}_{L})$, where $L <=> \space loaded$, $N <=> normal$.

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How to join it depends on how we choose the die. Are the two dice equally likely to be chosen? – vadim123 Apr 25 '13 at 14:03
The problem doesn't not say that they are equally likely, but if it is true, i think we should use the inclusion-exlusion principle. – blob Apr 25 '13 at 14:04
are you sure there are two faces with 1? – Alex Apr 25 '13 at 14:04
yes, the problem says there are two 1 – blob Apr 25 '13 at 14:05

Let $A$ be the event of choosing the regular die. Let $B$ be the event of choosing the loaded die. Let $C$ be the event of rolling a $1$. $Pr(C)=Pr(C\cap A)+Pr(C\cap B)$, because $A,B$ are complementary events. You have calculated $Pr(C|A)$ and $Pr(C|B)$ already. Combining, $Pr(C)=Pr(C|A)Pr(A)+Pr(C|B)Pr(B)$.