A box contains 90 discs, numbered from 1 to 90. If one disc is drawn at random from the box, the probability that it bears prime number less than 23 is
(a) $\frac{7}{90}$
(b) $\frac{1}{9}$
(c) $\frac{4}{15}$
(d) $\frac{8}{89}$
Total number of discs = 90.
Out of the given numbers, prime numbers less than 23 are 2, 3, 5, 7, 11, 13, 17 and 19.
Numbers of favourable outcomes = 8.
$\therefore P$ (getting a prime number which is less than 23 ) $=\frac{\text { Number of favourable outcomes }}{\text { Number of all possible outcomes }}$
$=\frac{8}{90}=\frac{4}{45}$
Thus, the probability that the disc bears prime number less than 23 is $\frac{4}{45}$.
Disclaimer: There is a misprinting in the question. If they ask for the probability of all prime numbers less than 90, then we get $\frac{4}{15}$.
Hence, the correct answer is option (c).
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