Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $L <=> \space loaded$, $N <=> normal$.

share|improve this question
    
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

1 Answer 1

up vote 1 down vote accepted

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, $Pr(C)=Pr(C|A)Pr(A)+Pr(C|B)Pr(B)$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.