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  • A box contains 4 while balls and 3 red balls. In succession, two balls are randomly selected and removed from the box. Given that the first removed ball is white, the probability that the second re

A box contains 4 while balls and 3 red balls. In succession, two balls are randomly selected and removed from the box. Given that the first removed ball is white, the probability that the second removed ball is red is

Option: 1

\frac{1}{3}


Option: 2

\frac{3}{7}


Option: 3

\frac{1}{2}


Option: 4

\frac{4}{7}


Answers (1)

best_answer

P(1st ball is white and 2nd is red),

                      \mathrm{p_1=\frac{4}{7} \times \frac{3}{6}=\frac{2}{7}}

                        

P(1st ball is white), \mathrm{p_2=\frac{4}{7}}

P(2nd will is red/1st ball is white)

                                 =\frac{\mathrm{\frac{p_1}{p_2}=\frac{2}{7}}}{\mathrm{\frac{4}{7}=\frac{1}{2}}}

Method II :

After removing Ist white ball, remaining balls in bag are 3 W & 3 R.

So P(red ball in 2 nd draw)

                                      =\frac{3}{6}=\frac{1}{2}

Posted by

Rishabh

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