91Ó°ÊÓ

Let X be the number of heads obtained when a coin is tossed.

The p.d.f of X is given by,

\({f_1}\left( x \right) = \left\{ {\begin{align}{}{6x\left( {1 - x} \right)}&{for\,0 < x < 1}\\0&{otherwise}\end{align}} \right.\)

Let C be the outcome of the selected coin.

Say\(C \in \left\{ {0,1} \right\}\)

Where,

\(C = 1\)Indicates that the outcome is H

\(C = 0\)Indicates that the outcome is T

The information given is that,

\(\begin{align}\Pr \left( {C = 1\left| {X = x} \right.} \right) = x\\\Pr \left( {C = 0\left| {X = x} \right.} \right) = 1 - x\end{align}\)

The joint pmf/pdf of X and C is given by

\(f\left( {x,y} \right) = \left\{ {\begin{align}{}{f\left( x \right)x}&{if}&{y = 1}\\{f\left( x \right)\left( {1 - x} \right)}&{if}&{y = 0}\end{align}} \right.\)

The conditional pdf of X given C=1, that is,

\(\begin{align}{g_2}\left( {x\left| {C = 1} \right.} \right) &= {g_2}\left( {x\left| {y = 1} \right.} \right)\\ &= \frac{{f\left( {x,1} \right)}}{{\Pr \left( {C = 1} \right)}}\end{align}\)

\(\begin{align}\Pr \left( {C = 1} \right) &= \int_0^1 {xf\left( x \right)dx} \\ &= \int_0^1 {6{x^2}\left( {1 - x} \right)dx} \\ &= 6\int_0^1 {\left( {{x^2} - {x^3}} \right)dx} \\ &= 6\left( {\frac{{{x^3}}}{3} - \frac{{{x^4}}}{4}} \right)_0^1\\ &= 6\left( {\frac{1}{3} - \frac{1}{4}} \right)\\ &= 6\left( {\frac{{4 - 3}}{{12}}} \right)\\ &= \frac{6}{{12}}\\ &= \frac{1}{2}\end{align}\)

The conditional pdf of X given that\(C = 1\)is,

\(\begin{align}\frac{{x.f\left( x \right)}}{{\Pr \left( {C = 1} \right)}} &= \frac{{6{x^2}\left( {1 - x} \right)}}{{\frac{1}{2}}}\\ &= 12{x^2}\left( {1 - x} \right),\,0 < x < 1\end{align}\)

Therefore, the conditional distribution of X is, \(12{x^2}\left( {1 - x} \right)\)