(a) A pair of dice is thrown simultaneously. If X denotes the absolute difference of numbers obtained on the pair of dice, then find the probability…

\multicolumn{5}{|l|}{(a)

$\displaystyle \mathrm{P}(\mathrm{X})$ & $\displaystyle \frac{6}{36}$ & $\displaystyle \frac{10}{36}$ & $\displaystyle \frac{8}{36}$ & $\displaystyle \frac{6}{36}$ & $\displaystyle \frac{4}{36}$ & $\displaystyle \frac{2}{36}$}

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$\displaystyle \mathrm{E}_{1}=$ Biased coin is selected $\displaystyle \Rightarrow \mathrm{P}\left(\mathrm{E}_{1}\right)=\frac{1}{2}$

$$\mathrm{E}_{$\displaystyle 2$}=\text { Fair coin is selected } \Rightarrow \mathrm{P}\left(\mathrm{E}_{$\displaystyle 2$}\right)=\frac{1}{2}

$$

A = Head appeared on tossing a selected coin .

$$\mathrm{P}\left(\frac{A}{\mathrm{E}_{$\displaystyle 1$}}\right)=\frac{1}{4}, \mathrm{P}\left(\frac{A}{\mathrm{E}_{$\displaystyle 2$}}\right)=\frac{1}{2}

$$By Bayes' Theorem $\displaystyle \mathrm{P}\left(\frac{\mathrm{E}_{1}}{A}\right)=\frac{\mathrm{P}\left(\mathrm{E}_{1}\right) \mathrm{P}\left(\dfrac{A}{\mathrm{E}_{1}}\right)}{\mathrm{P}\left(\mathrm{E}_{1}\right) \mathrm{P}\left(\dfrac{A}{\mathrm{E}_{1}}\right)+\mathrm{P}\left(\mathrm{E}_{2}\right) \mathrm{P}\left(\dfrac{A}{\mathrm{E}_{2}}\right)}$

$$\begin{gathered} =\frac{\dfrac{1}{2} \cdot \dfrac{1}{4}}{\dfrac{1}{2} \cdot \dfrac{1}{4}+\dfrac{1}{2} \cdot \dfrac{1}{2}} =\frac{1}{3} \end{gathered}