"If a random variable X follows the Binomial distribution B(5, p) such that P(X = 0) = P(X = 1), then ${{P(X = 2)} \over {P(X = 3)}}$ is equal to :"

Question

Accepted Answer

"

Given,

$n = 5$

and $P(X = 0) = P(X = 1)$

We know, $p(x = r) = {}^n{C_r}\,.\,{p^r}\,.\,{q^{n - r}}$

where $p + q = 1$

$\therefore$ $P(X = 0) = P(X = 1)$

$\Rightarrow {}^5{C_0}\,.\,{p^0}\,.\,{q^5} = {}^5{C_1}\,.\,{p^1}\,.\,{q^4}$

$\Rightarrow 1\,.\,1\,.\,{(1 - p)^5} = 5\,.\,p\,.\,{(1 - p)^4}$

$\Rightarrow 1 - p = 5p$

$\Rightarrow 6p = 1$

$\Rightarrow p = {1 \over 6}$

$\therefore$ $q = 1 - p = 1 - {1 \over 6} = {5 \over 6}$

Now, $${{P(X = 2)} \over {P(X = 3)}} = {{{}^5{C_2}\,.\,{p^2}\,.\,{q^3}} \over {{}^5{C_3}\,.\,{p^3}\,.\,{q^2}}}$$

$= {{10\,.\,q} \over {10\,.\,p}}$

$= {5 \over 6} \times {6 \over 1}$

$= 5$

"

If a random variable X follows the Binomial distribution B(5, p) such that P(X = 0) = P(X = 1), then ${{P(X = 2)} \over {P(X = 3)}}$ is equal to :