A die is thrown once ​​.Find the probabilty of getting
a)A doublet
b)A doublet of primes
c)A total of nine
d)An odd number on first die and a multiple of 3 on second die

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Please find below the solution to the asked query:

$$\begin{gathered} \mathrm{Assuming} \mathrm{that}, "\mathrm{A} \mathrm{die} \mathrm{is} \mathrm{thrown} \mathrm{twice}" \mathrm{not} \mathrm{once} \\ \mathrm{Total} \mathrm{cases}=6^2=36 \\ \left(\mathrm{i}\right) \mathrm{Doublet} \mathrm{will} \mathrm{be} \left\{\left(1,1\right),\left(2,2\right),\left(3,3\right),\left(4,4\right),\left(5,5\right),\left(6,6\right)\right\} \mathrm{i}.\mathrm{e}. 6 \mathrm{cases} \\ \mathrm{Required} \mathrm{probability}=\frac{6}{36}=\frac{1}{6} \\ \left(\mathrm{ii}\right) \mathrm{A} \mathrm{doublet} \mathrm{of} \mathrm{primes} \mathrm{can} \mathrm{be} \left\{\left(2,2\right),\left(3,3\right),\left(5,5\right)\right\} \\ \mathrm{Required} \mathrm{probability}=\frac{3}{36}=\frac{1}{12} \\ \left(\mathrm{iii}\right) \mathrm{An} \mathrm{odd} \mathrm{number} \mathrm{on} \mathrm{first} \mathrm{die} \mathrm{and} \mathrm{multiple} \mathrm{of} 3 \mathrm{on} \mathrm{other} \\ =\left\{\left(1,3\right),\left(1,6\right),\left(3,3\right),\left(3,6\right),\left(5,3\right),\left(5,6\right)\right\} \\ \mathrm{Required} \mathrm{probability}=\frac{6}{36}=\frac{1}{6} \end{gathered}$$

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