A box contains 3 white and 2 red balls. A bal | vedclass

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Question

(b) The second ball can be red in two different ways

$(i)$ First is white and second red

$P(A) = \frac{3}{5} 	imes \frac{2}{4} = \frac{6}{{20}}

Vedclass · Accepted Answer

$\frac{2}{5}$

Vedclass · Answer

(b) The second ball can be red in two different ways

$(i)$ First is white and second red

$P(A) = \frac{3}{5} 	imes \frac{2}{4} = \frac{6}{{20}}$

$(ii)$ First is red and second is also red

$P(B) = \frac{2}{5} 	imes \frac{1}{4} = \frac{2}{{20}}$

Both are mutually exclusive events,

hence required probability is $\frac{6}{{20}} + \frac{2}{{20}} = \frac{8}{{20}} = \frac{2}{5}.$

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

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