A) \[\frac{8}{25}\]
B) \[\frac{2}{5}\]
C) \[\frac{3}{5}\]
D) \[\frac{21}{25}\]
Correct Answer: B
Solution :
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 \[\frac{6}{20}+\frac{2}{20}=\frac{8}{20}=\frac{2}{5}.\]
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