Question

If $X$ and $Y$ are independent binomial variates $B\left( 5,\dfrac{1}{2} \right)$ and $B\left( 7,\dfrac{1}{2} \right)$, then find the value of $P\left( X+Y=3 \right)$. \[\]
A.$P\left( X+Y=3 \right)=\dfrac{45}{1024}$\[\]
B. $P\left( X+Y=3 \right)=\dfrac{65}{1024}$\[\]
C. $P\left( X+Y=3 \right)=\dfrac{55}{1024}$\[\]
D. $P\left( X+Y=3 \right)=\dfrac{75}{1024}$\[\]

Answer 1

Hint: We recall the joint probability mass function for binomial variates $P\left( X+Y=k \right)=\sum\limits_{a,b}{{}^{n}{{C}_{a}}{{p}^{n}}\cdot {}^{m}{{C}_{b}}{{q}^{m}}}$ where $n,m$ are the number of trials, $p,q$ are the probabilities of successes of a single trial, $a,b$ are the number of successes of the random variables $X$ and $Y$ respectively and $k=a+b$ is the sum of number of successes. We can get $k=3$ successes if $a=0,b=3$ or $a=2,b=1$ or $a=1,b=2$ or $a=3,b=0$. \[\]

Complete step by step answer:
We know that if random variable $X$ follows binomial distribution $\left( X\tilde{\ }B\left( n,p \right) \right)$ with number of trials $n\in \mathsf{\mathbb{N}}$ and probability of success $p\in \left[ 0,1 \right]$the probability that we get $k$ successes in $n$ independent trials is given by the probability mass function,
\[P\left( X=k \right)={}^{n}{{C}_{k}}{{p}^{k}}{{\left( 1-p \right)}^{n-k}}\]
Let us consider two independent random variables $X$ and $Y$ follow binomial distribution with $n$ and $m$ trials with probability of success in one trial $p$ and $q$ then the probability mass function for total $k$ successes with $X$ and $Y$ combined is given by ;
\[P\left( X+Y=k \right)=\sum\limits_{a,b}{{}^{n}{{C}_{a}}{{p}^{n}}\cdot {}^{m}{{C}_{b}}{{q}^{m}}}\]
Here $a$ is the number of successes of $X$ within $n$ trials and $b$ is the number of successes of $Y$ with $m$ trials which means $a+b=k$\[\]
We are given the question that $X$ and $Y$ are independent binomial variates with binomial distribution. $B\left( 5,\dfrac{1}{2} \right)$ and $B\left( 7,\dfrac{1}{2} \right)$ . So the number of trials for $X$ is $n=5$ and for $Y$ is $m=7$. The probability of success of a single trial is $p=q=\dfrac{1}{2}$. We are asked to find $P\left( X+Y=3 \right)$ which means we are asked for the probability we get $k=3$ successes in the separate(independent ) trials for $X$ and $Y$. \[\]
If we get $a=0$ success for $X$ and $b=3$ successes for $Y$ then the probability for $a+b=3+0=3=k$ successes is;
\[P\left( X=0,Y=3 \right)={}^{5}{{C}_{0}}{{\left( \dfrac{1}{2} \right)}^{5}}{}^{7}{{C}_{3}}{{\left( \dfrac{1}{2} \right)}^{7}}=1\times \dfrac{1}{32}\times 35\times \dfrac{1}{32}=\dfrac{35}{4096}\]
If we get $a=1$ success for $X$ and $b=2$ successes for $Y$ then the probability for $a+b=1+2=3=k$ successes is;
\[P\left( X=1,Y=2 \right)={}^{5}{{C}_{1}}{{\left( \dfrac{1}{2} \right)}^{5}}{}^{7}{{C}_{2}}{{\left( \dfrac{1}{2} \right)}^{7}}=5\times \dfrac{1}{32}\times 21\times \dfrac{1}{32}=\dfrac{105}{4096}\]
If we get $a=2$ success for $X$ and $b=1$ successes for $Y$ then the probability for $a+b=2+1=3=k$ successes is;
\[P\left( X=2,Y=1 \right)={}^{5}{{C}_{2}}{{\left( \dfrac{1}{2} \right)}^{5}}{}^{7}{{C}_{1}}{{\left( \dfrac{1}{2} \right)}^{7}}=10\times \dfrac{1}{32}\times 7\times \dfrac{1}{32}=\dfrac{70}{4096}\]
If we get $a=3$ success for $X$ and $b=0$ successes for $Y$ then the probability for $a+b=0+3=3=k$ successes is;
\[P\left( X=3,Y=0 \right)={}^{5}{{C}_{3}}{{\left( \dfrac{1}{2} \right)}^{5}}{}^{7}{{C}_{0}}{{\left( \dfrac{1}{2} \right)}^{7}}=5\times \dfrac{1}{32}\times 1\times \dfrac{1}{32}=\dfrac{10}{4096}\]
So the probability that we get $k=3$ successes in independent bivariate $X,Y$ is ;
\[\begin{align}
  & P\left( X+Y=3 \right)=P\left( X=0,Y=3 \right)+P\left( X=0,Y=3 \right)+P\left( X=0,Y=3 \right)+P\left( X=0,Y=3 \right) \\
 & \Rightarrow P\left( X+Y=3 \right)=\dfrac{35}{4096}+\dfrac{105}{4096}+\dfrac{70}{4096}+\dfrac{10}{4096}=\dfrac{220}{4096}=\dfrac{55}{1024} \\
\end{align}\]

So, the correct answer is “Option C”.

Note: We note that binomial distribution is a discrete probability distribution which is defined for trials which returns a Boolean valued outcome. If $X$ and $Y$ are random variables then $X+Y$ is also a random variable. We can alternatively solve using the formula $P\left( X+Y=k \right)={}^{n+m}{{C}_{k}}{{p}^{k}}{{\left( 1-p \right)}^{n+m-k}}$ . We can two random variables $X,Y$ independent if $P\left( X=k,Y=l \right)=P\left( X=k \right)P\left( Y=l \right)$ .