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Let $X_1$ and $X_2$ be independent random variables with respective binomial distributions $b(3,1/2)$ and $b(5,1/2)$. Find $P(X_1+X_2=7)$.

I am thinking of an example I pulled from my notes: Ex: let $X_1$ ~ Poisson $(\lambda_1=2)$ and $X_2$~Poisson$(\lambda_2=3)$ be independent, then $P(X_1+X_2=1)=P(X_1=0,X_2=1)+P(X_1=1,X_2=0)=(\lambda_1+\lambda_2)e^{-(\lambda_1+\lambda_2)}$

So my thoughts for this question: $P(x,p)=\binom{n}{x}(p)^x (1-p)^{n-x}$. So for this question, should I cconsider by cases, like $X_1=1, X_2=3$ etc, and since they are independent, then I would add those cases together?

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    $\begingroup$ Think what is the distribution of $X_1+X_2$, you need not formulas for it. $\endgroup$ – kludg Mar 10 '20 at 3:02
  • $\begingroup$ can you specify? @kludg $\endgroup$ – shine Mar 10 '20 at 3:08
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The binomial distribution with amount $n$ and success rate $p$, is more often denoted $\mathcal{Bin}(n,p)$.

$Z\sim\mathcal{Bin}(n,p)$ denotes that random variable, $Z$ is a count of successes among $n$ independent Bernoulli trials with identical success rate $p$.

So given that the independent random variables are $X_1\sim\mathcal{Bin}(3, 1/2)$ and $X_2\sim\mathcal{Bin}(5,1/2)$, what does $X_1+X_2$ count?   What is the distribution for this?

$X_1$ counts successes among $3$ independent Bernoulli trials with identical success rate $1/2$, and independently $X_2$ counts successes among $7$ independent Bernoulli trials with identical success rate $1/2$.

So $X_1+X_2$ counts...

Use this.


Okay, it is possible to use the Law of Total Probability.

Now, when the sum is $7$ the minimum $X_1$ may be is $2$ (because the maximum $X_2$ may be is $5$), so …

$$\begin{align}\mathsf P(X_1+X_2=7)~&=~\sum_{k=2}^3\mathsf P(X_1=k)~\mathsf P(X_2=7-k)\\[1ex]&=~\mathsf P(X_1=2)\,\mathsf P(X_2=5)+\mathsf P(X_1=3)\,\mathsf P(X_2=4)\\[1ex]&~~\vdots\end{align}$$

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  • $\begingroup$ I am still not sure of this , can you specify even more on counting $X_1+X_2$ and its distribution? $\endgroup$ – shine Mar 10 '20 at 3:23
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    $\begingroup$ @shine … No. .. $\endgroup$ – Graham Kemp Mar 10 '20 at 3:47

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