A bag contains 3 black and 4 white balls. Two balls are drawn one by one at random without replaceme

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14.Probability

hard

A bag contains $3$ black and $4$ white balls. Two balls are drawn one by one at random without replacement. The probability that the second drawn ball is white, is

A

$\frac{4}{{49}}$

B

$\frac{1}{7}$

C

$\frac{4}{7}$

D

$\frac{{12}}{{49}}$

Solution

(c) Second white ball can draw in two ways.

$(i)$ First is white and second is white

Probability $ = \frac{4}{7} \times \frac{3}{6} = \frac{2}{7}$

$(ii)$ First is black and second is white

Probability $ = \frac{3}{7} \times \frac{4}{6} = \frac{2}{7}$

Hence required probability $ = \frac{2}{7} + \frac{2}{7} = \frac{4}{7}.$

Standard 11

Mathematics

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