$\begin {array} {1 1} (1)\;\large\frac{17}{56} & \quad (2)\;\large\frac{12}{95} \\ (3)\;\large\frac{17}{95} & \quad (4)\;None\: of \: these \end {array}$

The urn consideration 10 blue and 6 red balls.

P( exactly 3 red balls ) = $ \large\frac{6C_3}{16C_3}$

P( exactly 2 red balls ) = $ \large\frac{6C_2 \times 10C_1}{16C_3}$

$ \therefore $ P( at least 2 red balls ) = $ \large\frac{6C_3}{16C_3} + \large\frac{6C_2 \times 10C_1}{16C_3}$

$ = \large\frac{20+15 \times 1\not{0}}{56\not{0}} = \large\frac{20+150}{560}$

$ \large\frac{17}{56}$

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