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

Hint: 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.