$\begin {array} {1 1} (A)\;\large\frac{3}{5} & \quad (B)\;\large\frac{5}{6} \\ (C)\;\large\frac{1}{36} & \quad (D)\;None\: of \: these \end {array}$

Given :

Three mangoes

Three apples

$ \therefore$ Total fruits = 6

$ \Rightarrow$ Out of 6 fruits 2 fruits can be selected in $ ^6 C_2$ ways

Also 1 apple and 1 mango can be selected in $ ^3C_1 \times ^3C_1=3\times 3 $ ways

Hence the required probability is $ \large\frac{9}{^6C_2}$

$ = \large\frac{9}{15} = \large\frac{3}{5}$

Ans : (A)

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