Question

Let the random variable $X$ follow $B\left( {6,p} \right)$ . If $16P\left( {X = 4} \right) = P\left( {X = 2} \right)$ , then what is the value of $p$?
A.$\dfrac{1}{3}$
B.$\dfrac{1}{4}$
C.$\dfrac{1}{5}$
D.$\dfrac{1}{6}$

Answer 1

Here we need to find the value of the variable. We will use the formula of binomial probability and expand the given probability individually. Then we will substitute all the given values and simplify it. We will then back substitute the obtained values in the given expression and simplify it further using a combination formula to get the required value.

Formula used:
We will use the formula for binomial probability which is given by $P\left( {X = t} \right) = {}^n{C_t}{p^t}{\left( {1 - p} \right)^{n - t}}$.

Complete step-by-step answer:
It is given that $X$ follow $B\left( {6,p} \right)$, where $X$ is a random variable.
Here, $B\left( {6,p} \right)$ means $n = 6$ and $p = {\rm{probability}}$
We will first find the value of $P\left( {X = 4} \right)$.
By substituting $n = 6$ and $t = 4$ in the formula $P\left( {X = t} \right) = {}^n{C_t}{p^t}{\left( {1 - p} \right)^{n - t}}$, we get
$\Rightarrow P\left( {X = 4} \right) = {}^6{C_4}{p^4}{\left( {1 - p} \right)^{6 - 4}}$
On further simplification, we get
$\Rightarrow P\left( {X = 4} \right) = {}^6{C_4}{p^4}{\left( {1 - p} \right)^2}$ ……….. $\left( 1 \right)$
Similarly, we will find the value of $P\left( {X = 2} \right)$.
By substituting $n = 6$ and $t = 2$ in the formula $P\left( {X = t} \right) = {}^n{C_t}{p^t}{\left( {1 - p} \right)^{n - t}}$, we get

$P\left( {X = 2} \right) = {}^6{C_2}{p^2}{\left( {1 - p} \right)^{6 - 2}}$
On further simplification, we get
$\Rightarrow P\left( {X = 2} \right) = {}^6{C_2}{p^2}{\left( {1 - p} \right)^4}$ ……….. $\left( 2 \right)$
The given expression is $16P\left( {X = 4} \right) = P\left( {X = 2} \right)$.
Now, we will substitute the values obtained in equation $\left( 1 \right)$ and equation $\left( 2 \right)$ here in this given expression.
$\Rightarrow 16{}^6{C_4}{p^4}{\left( {1 - p} \right)^2} = {}^6{C_2}{p^2}{\left( {1 - p} \right)^4}$
Dividing both sides by ${p^2}$, we get
$\Rightarrow 16{}^6{C_4}{p^2} = {}^6{C_2}{\left( {1 - p} \right)^2}$
Using this formula of combination ${}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$, we get
$16 \times \dfrac{{6!}}{{4!\left( {6 - 4} \right)!}} \times {p^2} = \dfrac{{6!}}{{2!\left( {6 - 2} \right)}} \times {\left( {1 - p} \right)^2}$
On subtracting the terms inside the bracket, we get
$\Rightarrow 16 \times \dfrac{{6!}}{{2! \times 4!}} \times {p^2} = \dfrac{{6!}}{{4! \times 2!}} \times {\left( {1 - p} \right)^2}$
On further simplification, we get
$\Rightarrow 16 \times {p^2} = {\left( {1 - p} \right)^2}$
Rewriting the expression, we get
$\Rightarrow 16{p^2} - {\left( {1 - p} \right)^2} = 0$
Using algebraic identities ${a^2} - {b^2} = \left( {a - b} \right)\left( {a + b} \right)$, we get
$\Rightarrow \left( {4p - 1 + p} \right)\left( {4p + 1 - p} \right) = 0$
On further simplifying the terms, we get
$\Rightarrow \left( {5p - 1} \right)\left( {3p + 1} \right) = 0$
Using the zero product property, we get
$\Rightarrow 3p + 1 = 0$ or $5p - 1 = 0$
Simplifying the expression, we get
$\Rightarrow p = - \dfrac{1}{3}$ or $p = \dfrac{1}{5}$
But we know that the probability can’t be negative.
Therefore, $p = \dfrac{1}{5}$
Hence, the correct option is option C.

Note: Here we have obtained the probability $p$, so we need to know about probability and its properties. Probability is defined as the ratio of favorable outcomes to the total number of outcomes. We need to keep in mind that the value of probability cannot be greater than 1 and the value of probability cannot be negative. Also, the probability of a sure event is always one, however, if the probability of the event is 0, then the event will never occur.