$\begin {array} {1 1} (A)\;\large\frac{1}{5} & \quad (B)\;\large\frac{4}{5} \\ (C)\;\large\frac{3}{5} & \quad (D)\;\large\frac{2}{5} \end {array}$

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$E_1$ can occur as (1,5) (5,1) (2,4) (4,2) (3,3)

$E_2$ can happen as

(2,1) (2,2) (2,3) (2,4) (2,5) (2,6) (6,2) (5,2) (4,2) (3,2) (1,2)

$ \therefore P(E_2/E_1) = $ Probability of $E_2$ when $E_1$ has occured

$ = \large\frac{2}{5} \because (2,4)\: and \: (4,2)$ are the cases out of five.

Alternatively $ P( E_2/ E_1) = \large\frac{P(E_1 \cap E_2)}{P(E_1)}$

$ = \Large\frac{\Large\frac{2}{36}}{\Large\frac{5}{36}}$

$ = \large\frac{2}{\not{36}} \times \large\frac{\not{36}}{5}$

$ = \large\frac{2}{5}$

Ans : (D)

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