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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
