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14.Probability
hard
A box contains $3$ white and $2$ red balls. A ball is drawn and another ball is drawn without replacing first ball, then the probability of second ball to be red is
A
$\frac{8}{{25}}$
B
$\frac{2}{5}$
C
$\frac{3}{5}$
D
$\frac{{21}}{{25}}$
Solution
(b) The second ball can be red in two different ways
$(i)$ First is white and second red
$P(A) = \frac{3}{5} \times \frac{2}{4} = \frac{6}{{20}}$
$(ii)$ First is red and second is also red
$P(B) = \frac{2}{5} \times \frac{1}{4} = \frac{2}{{20}}$
Both are mutually exclusive events,
hence required probability is $\frac{6}{{20}} + \frac{2}{{20}} = \frac{8}{{20}} = \frac{2}{5}.$
Standard 11
Mathematics
