A die is loaded such that an even number is likely to come thrice as likely as odd number.the dice is thrown twice. Find the probability distribution of random variable represent as the prime number on 2 tosses

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$$\begin{gathered} \mathrm{We} \mathrm{have} \\ \mathrm{A} \mathrm{die} \mathrm{is} \mathrm{loaded} \mathrm{such} \mathrm{that} \mathrm{an} \mathrm{even} \mathrm{number} \mathrm{is} \mathrm{likely} \mathrm{to} \mathrm{come} \mathrm{thrice} \mathrm{as} \mathrm{likely} \mathrm{as} \mathrm{odd} \mathrm{number}. \\ \mathrm{If} \mathrm{E} \mathrm{represents} \mathrm{even} \mathrm{number} \mathrm{and} \mathrm{O} \mathrm{represents} \mathrm{odd}, \mathrm{then} \\ \mathrm{P}\left(\mathrm{E}\right)=3\mathrm{P}\left(\mathrm{O}\right) \\ \mathrm{When} \mathrm{a} \mathrm{die} \mathrm{is} \mathrm{loaded}, \mathrm{we} \mathrm{get} 1 \mathrm{or} 2 \mathrm{or} 3 \mathrm{or} 4 \mathrm{or} 5 \mathrm{or} 6 \\ \mathrm{P}\left(1\right)+\mathrm{P}\left(2\right)+\mathrm{P}\left(3\right)+\mathrm{P}\left(4\right)+\mathrm{P}\left(5\right)+\mathrm{P}\left(6\right)=1 \\ \Rightarrow \mathrm{x}+3\mathrm{x}+\mathrm{x}+3\mathrm{x}+\mathrm{x}+3\mathrm{x}=1 \\ \Rightarrow \mathrm{x}=\frac{1}{12} \\ \mathrm{Now} \mathrm{prime} \mathrm{numbers} \mathrm{re} 2,3,5 \\ \mathrm{P}\left(2\right)=3\mathrm{x}=\frac{1}{4} \\ \mathrm{P}\left(3\right)=\mathrm{P}\left(5\right)=\frac{1}{12} \\ \mathrm{Now} \\ \mathrm{Dice} \mathrm{is} \mathrm{rolled} \mathrm{twice}, \mathrm{so} \\ \mathrm{we} \mathrm{may} \mathrm{get} \mathrm{no} \mathrm{prime} \mathrm{no}., 1 \mathrm{prime} \mathrm{number} \mathrm{or} 2 \mathrm{prime} \mathrm{number} \\ \mathrm{If} \mathrm{X} \mathrm{represents} \mathrm{number} \mathrm{of} \mathrm{prime} \mathrm{number}, \mathrm{then} \\ \mathrm{P}\left(\mathrm{X}=1\right)= 2 \mathrm{or} 3 \mathrm{or} 5=\frac{1}{4}+\frac{1}{12}+\frac{1}{12}=\frac{4}{12} \\ \mathrm{P}\left(\mathrm{X}=2\right)=\left(2,3\right) \mathrm{or} \left(3,2\right) \mathrm{or}\left(3,5\right) \mathrm{or} \left(5,3\right) \mathrm{or} \left(2,5\right) \mathrm{or} \left(5,2\right) \\ =\frac{1}{48}+\frac{1}{48}+\frac{1}{144}+\frac{1}{144}+\frac{1}{48}+\frac{1}{48} \\ =\frac{14}{144} \\ =\frac{7}{72} \\ \mathrm{P}\left(\mathrm{X}=0\right)=1-\left(\mathrm{P}\left(\mathrm{X}=1\right) +\mathrm{P}\left(\mathrm{X}=2\right)\right) \\ =1-\left[\frac{4}{12}+\frac{7}{72}\right] \\ =1-\left[\frac{24}{72}+\frac{7}{72}\right] \\ =\frac{41}{72} \end{gathered}$$

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