A) \[\frac{6}{25}\]
B) \[\frac{1}{4}\]
C) \[\frac{1}{6}\]
D) \[\frac{2}{5}\]
E) \[\frac{4}{5}\]
Correct Answer: A
Solution :
E = set of number divisible by 6 \[=\{6,12,18,\text{ }24,\text{ }30,...,96\}\] F = set of number divisible by 8 \[=\{8,16,\text{ }24,...,96\}\] \[E\cap F=\]set of number divisible by 6 and 8 \[=\{24,\text{ }48,\text{ }72,\text{ }96\}\] \[\therefore \]\[n(E)=16,n(F)=12,n(E\cap F)=4\] \[\therefore \]\[n(E\cup F)=n(E)+n(F)+n(F)-n(E\cap F)\] \[=16+12-4=24\] and \[n(S)=100.\] \[\therefore \]Required probability \[\frac{n(E\cup F)}{n(S)}=\frac{24}{100}=\frac{6}{25}\]
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