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Home  >>  JEEMAIN and AIPMT  >>  Mathematics  >>  Class12  >>  Probability
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A bag contains 3 black, 3 white and 2 red balls. One by one three balls are drawn without replacement. The probability that the third ball is red is equal to

$\begin {array} {1 1} (A)\;\large\frac{3}{56} & \quad (B)\;\large\frac{1}{56} \\ (C)\;\large\frac{12}{56} & \quad (D)\\large\frac{14}{56} \end {array}$

 

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1 Answer

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Three possible cases are
$Ist \: draw\: \: \: \: \: IInd \: draw\: \: \: \: \: IIIrd\: draw$
$ Red\: \: \: \: \: \: \: \: \: \: \: \: \: \: Non\: Red\: \: \: \: \: \: \: \: Red$
$Non\: Red\: \: \: \: \: \: \: \: \: \: \: Red\:\: \: \: \:  \: \: \: \: \: \: Red$
$ Non\: Red\: \: \: \: \:\: \: \: \: Non\: Red\: \: \: \: \: Red$
$ \therefore $ Required probability = $ \large\frac{2}{8} \times \large\frac{6}{7} \times \large\frac{1}{6}+ \large\frac{6}{8} \times \large\frac{2}{7} \times \large\frac{1}{6}+ \large\frac{6}{8} \times \large\frac{5}{7} \times \large\frac{2}{6}$
$ = \large\frac{14}{56}$
Ans : (D)

 

answered Jan 16, 2014 by thanvigandhi_1
 

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