Question

If X and Y are the independent binomial variates $B\left( {5,\dfrac{1}{2}} \right)$ and$B\left( {7,\dfrac{1}{2}} \right)$, then $P\left( {X + Y = 3} \right)$ is equal to
A. $\dfrac{{35}}{{47}}$
B. $\dfrac{{55}}{{1024}}$
C. $\dfrac{{220}}{{512}}$
D. $\dfrac{{11}}{{204}}$

Answer 1

From the given we are asked to find the value$P\left( {X + Y = 3} \right)$. For that, we will first find all the possible values of $X$and $Y$ from the equation $X + Y = 3$then we will find the values $P\left( {X = r} \right)$and$P\left( {Y = r} \right)$. Then we will make use of these to find the value of$P\left( {X + Y = 3} \right)$.

Formula Used: Some formulas that we need to know before solving this problem:
$^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$
$^n{C_0} = 1$

Complete step-by-step solution:
It is given that $X$ and $Y$ are the two independent binomial variates $B\left( {5,\dfrac{1}{2}} \right)$ and$B\left( {7,\dfrac{1}{2}} \right)$. We aim to find the value of$P\left( {X + Y = 3} \right)$. First, we need to find the possible values of X and Y from the equation$X + Y = 3$.
Consider the equation$X + Y = 3$, if $X = 0$then$Y = 3$.
If $X = 1$then$Y = 2$.
If $X = 2$then$Y = 1$.
If $X = 3$then$Y = 0$.
Now we got all the possible values of X and Y. Let us find $P\left( {X = r} \right)$and$P\left( {Y = r} \right)$.
Here it is given that $B\left( {5,\dfrac{1}{2}} \right)$ and$B\left( {7,\dfrac{1}{2}} \right)$. Thus, we get
P\left( {X = r} \right) = $$$$^5{C_r}{\left( {\dfrac{1}{2}} \right)^r}{\left( {\dfrac{1}{2}} \right)^{5 - r}} = $$$$^5{C_r}{\left( {\dfrac{1}{2}} \right)^5}………….$(1)$
P\left( {Y = r} \right) = $$$$^7{C_r}{\left( {\dfrac{1}{2}} \right)^7}…………….$(2)$
Now let us find the value of $P\left( {X + Y = 3} \right)$
$P\left( {X + Y = 3} \right) = P\left( {X = 0} \right)P\left( {Y = 3} \right) + P\left( {X = 1} \right)P\left( {Y = 2} \right) + P\left( {X = 2} \right)P\left( {Y = 1} \right) + P\left( {X = 3} \right)P\left( {Y = 0} \right)$
Using the equations $\left( 1 \right)$and$(2)$ the above expression can be written as
= $$$$^5{C_0}{\left( {\dfrac{1}{2}} \right)^5}{.^7}{C_3}{\left( {\dfrac{1}{2}} \right)^7} + {{\text{ }}^5}{C_1}{\left( {\dfrac{1}{2}} \right)^5}{.^7}{C_2}{\left( {\dfrac{1}{2}} \right)^7} + {{\text{ }}^5}{C_2}{\left( {\dfrac{1}{2}} \right)^5}{.^7}{C_1}{\left( {\dfrac{1}{2}} \right)^7} + {{\text{ }}^5}{C_3}{\left( {\dfrac{1}{2}} \right)^5}{.^7}{C_0}{\left( {\dfrac{1}{2}} \right)^7}
Let us take ${\left( {\dfrac{1}{2}} \right)^{12}}$commonly out from the above expression.
$= {\left( {\dfrac{1}{2}} \right)^{12}}\left[ {^5{C_0}{.^7}{C_3} + {{\text{ }}^5}{C_1}{.^7}{C_2} + {{\text{ }}^5}{C_2}{.^7}{C_1} + {{\text{ }}^5}{C_3}{.^7}{C_0}} \right]$
On simplifying this using, the formulas $^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$and $^n{C_0} = 1$we get
$= {\left( {\dfrac{1}{2}} \right)^{12}}\left[ {1 \times \dfrac{{7 \times 6 \times 5}}{6} + 5 \times \dfrac{{7 \times 6}}{2} + \dfrac{{5 \times 4}}{2} \times 7 + \dfrac{{5 \times 4}}{2} \times 1} \right]$
On simplifying it further we get
$Y$
Let us simplify it even more
$= \dfrac{1}{{{2^{12}}}} \times 220$
$= \dfrac{1}{{{2^{12}}}} \times 220$
$= \dfrac{{55}}{{{2^{10}}}}$
$P\left( {X + Y = 3} \right) = \dfrac{{55}}{{1024}}$
Thus, we have found the value of$P\left( {X + Y = 3} \right)$. Now let us see the options to find the right answer.
Option (a) $\dfrac{{35}}{{47}}$is an incorrect answer as we got $P\left( {X + Y = 3} \right) = \dfrac{{55}}{{1024}}$from our calculation.
Option (b) $\dfrac{{55}}{{1024}}$is the correct answer as we got the same value in our calculation above.
Option (c) $\dfrac{{220}}{{512}}$is an incorrect answer as we got $P\left( {X + Y = 3} \right) = \dfrac{{55}}{{1024}}$from our calculation.
Option (d) $\dfrac{{11}}{{204}}$is an incorrect answer as we got $P\left( {X + Y = 3} \right) = \dfrac{{55}}{{1024}}$from our calculation.
Hence, option (b) $\dfrac{{55}}{{1024}}$is the correct answer.

Note: In this problem, it is necessary to find the possible values of X and Y as we want them to find the value of$P\left( {X + Y = 3} \right)$. We can also find the values$P\left( {X + Y = 3} \right) = P\left( {X = 0} \right)P\left( {Y = 3} \right) + P\left( {X = 1} \right)P\left( {Y = 2} \right) + P\left( {X = 2} \right)P\left( {Y = 1} \right) + P\left( {X = 3} \right)P\left( {Y = 0} \right)$ one by one separately to make our calculation easier.