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Home  >>  JEEMAIN and AIPMT  >>  Mathematics  >>  Class12  >>  Probability
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A box contains 3 white and 2 red balls. If we draw one ball and without replacing the first ball, the probability of drawing red ball in the second draw is

$\begin {array} {1 1} (A)\;\large\frac{8}{25} & \quad (B)\;\large\frac{2}{5} \\ (C)\;\large\frac{3}{5} & \quad (D)\;\large\frac{21}{25} \end {array}$


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Case (i) The first ball is white and second is red.
Its probability = $ \large\frac{3}{5} \times \large\frac{2}{4}$
$ = \large\frac{3}{10}$
Case (ii) The first ball is red and second is red.
Its probability = $ \large\frac{ \not 2}{5} \times \large\frac{1}{\not 4\: 2}$
$ = \large\frac{1}{10}$
$ \therefore $ Required probability = $ \large\frac{3}{10} +\large\frac{1}{10}$
$ = \large\frac{4}{10} = \large\frac{2}{5}$
Ans : (B)
answered Jan 8, 2014 by thanvigandhi_1

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