Question

From a set of 100 cards numbered 1 to 100, one card is drawn at random. The probability that the number obtained on the card is divisible by 6 or 8 but not by 24, is:

• A. $ \frac{6}{25} $
• B. $ \frac{1}{4} $
• C. $ \frac{1}{6} $
• D. $ \frac{2}{5} $
• E. $ \frac{4}{5} $

Answer 1

Answer: (A)

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} $