Question

Let a random variable X have a binomial distribution with mean 8 and variance 4. If \[P(x\le 2)=\dfrac{k}{{{2}^{16}}}\], then k is equal to?
(A) 17
(B) 1
(C) 121
(D) 137

Answer 1

Hint: We solve this problem by considering our random variable as Binomial with parameters n, p and we use the formula that $Mean=np$ and $Variance=npq$ and solve them to find the values of n, p and q. Then we use the formula for probability \[P(X=x)={}^{n}{{C}_{x}}{{p}^{x}}{{q}^{n-x}}\] and find the probability for all x less than or equal to 2 and compare the obtained result with given value and find the required value.

Complete step-by-step answer:
We are given that the random variable X follows the Binomial Distribution with mean 8 and variance 4.
Now let us consider the concept of Binomial Distribution.
A random variable X is said to follow Binomial distribution if its probability can be given as,
$P\left( X=x \right)={}^{n}{{C}_{x}}{{p}^{x}}{{q}^{n-x}}\text{ }q=1-p\text{ for }x=1,2,3,......,n$
If X follows a Binomial distribution with parameters n and p, then mean and variance of X are
$\begin{align}
  & Mean=np \\
 & Variance=npq \\
\end{align}$
So, using this formula we can write as
\[np=8..........(1)\]
\[npq=4............(2)\]
Now, let us divide equation (2) with equation (1). Then we get,
\[\begin{align}
  & \Rightarrow \dfrac{npq}{np}=\dfrac{4}{8} \\
 & \Rightarrow q=\dfrac{1}{2} \\
\end{align}\]
From the definition of Binomial distribution, we can write,
\[\Rightarrow p+q=1\]
\[\Rightarrow p=1-q\]
\[\Rightarrow p=1-\dfrac{1}{2}\]
\[\Rightarrow p=\dfrac{1}{2}\]
So, our random variable X has parameters as \[p=\dfrac{1}{2}\] and \[q=\dfrac{1}{2}\].
Now let us substitute these values in equation (1). Then we get,
\[\begin{align}
  & \Rightarrow np=8 \\
 & \Rightarrow n\left( \dfrac{1}{2} \right)=8 \\
 & \Rightarrow n=8\times 2 \\
 & \Rightarrow n=16 \\
\end{align}\]
We know,
\[P(x\le 2)=P(x=0)+P(x=1)+P(x=2)\]
As we know, the probability of a Binomial Distribution at \[X=x\] is denoted by \[P(X=x)\]is
\[P(X=x)={}^{n}{{C}_{x}}{{p}^{x}}{{q}^{n-x}}\]
\[\Rightarrow P(x\le 2)=P(x=0)+P(x=1)+P(x=2)\]
\[\Rightarrow P(x\le 2)={}^{16}{{C}_{0}}{{p}^{0}}{{q}^{16-0}}+{}^{16}{{C}_{1}}{{p}^{1}}{{q}^{16-1}}+{}^{16}{{C}_{2}}{{p}^{2}}{{q}^{16-2}}\]
\[\Rightarrow P(x\le 2)={}^{16}{{C}_{0}}{{\left( \dfrac{1}{2} \right)}^{0}}{{\left( \dfrac{1}{2} \right)}^{16}}+{}^{16}{{C}_{1}}{{\left( \dfrac{1}{2} \right)}^{1}}{{\left( \dfrac{1}{2} \right)}^{15}}+{}^{16}{{C}_{2}}{{\left( \dfrac{1}{2} \right)}^{2}}{{\left( \dfrac{1}{2} \right)}^{14}}\]
\[\Rightarrow P(x\le 2)={}^{16}{{C}_{0}}{{\left( \dfrac{1}{2} \right)}^{16}}+{}^{16}{{C}_{1}}{{\left( \dfrac{1}{2} \right)}^{16}}+{}^{16}{{C}_{2}}{{\left( \dfrac{1}{2} \right)}^{16}}\]
\[\Rightarrow P(x\le 2)={{\left( \dfrac{1}{2} \right)}^{16}}\left( {}^{16}{{C}_{0}}+{}^{16}{{C}_{1}}+{}^{16}{{C}_{2}} \right)\]
We know
\[{}^{n}{{C}_{x}}=\dfrac{n!}{(n-r)!r!}\]
\[\Rightarrow P(x\le 2)={{\left( \dfrac{1}{2} \right)}^{16}}\left( \dfrac{16!}{(16-0)!0!}+\dfrac{16!}{(16-1)!1!}+\dfrac{16!}{(16-2)!2!} \right)\]
\[\Rightarrow P(x\le 2)=\left( \dfrac{1}{{{2}^{16}}} \right)\left( \dfrac{16!}{16!}+\dfrac{16!}{15!1!}+\dfrac{16!}{14!2!} \right)\]
\[\Rightarrow P(x\le 2)=\left( \dfrac{1}{{{2}^{16}}} \right)\left( 1+16+\dfrac{16\times 15}{2} \right)\]
\[\Rightarrow P(x\le 2)=\left( \dfrac{1}{{{2}^{16}}} \right)\left( 1+16+8\times 15 \right)\]
\[\Rightarrow P(x\le 2)=\left( \dfrac{1}{{{2}^{16}}} \right)\left( 1+16+120 \right)\]
\[\Rightarrow P(x\le 2)=\left( \dfrac{1}{{{2}^{16}}} \right)\left( 137 \right)\]
We are given that \[P(x\le 2)=\dfrac{k}{{{2}^{16}}}\]. So, let us compare both the values. Then we get,
\[\left( \dfrac{1}{{{2}^{16}}} \right)\left( 137 \right)=\dfrac{k}{{{2}^{16}}}\]
\[\Rightarrow k=137\]
Hence the value of k is 137.
Therefore, Option D is correct.

Note: There is a possibility of one making a mistake by taking the probability of a binomial distribution as \[P(X=x)={}^{n}{{P}_{x}}{{p}^{x}}{{q}^{n-x}}\]. But the formula is \[P(X=x)={}^{n}{{C}_{x}}{{p}^{x}}{{q}^{n-x}}\]. Here \[{}^{n}{{C}_{x}}=\dfrac{n!}{(n-r)!r!}\] and \[{}^{n}{{P}_{x}}=\dfrac{n!}{(n-r)!}\]. So, one should remember the difference between permutations and combinations.