A box contains 90 discs, numbered from 1 to 90. If one disk is drawn at random from the box, the probability that it bears a prime number less than 23, is:
$( A)\frac{7}{90}$
$( B)\frac{10}{90}$
$( C)\frac{4}{45}$
$( D)\frac{9}{89}$
Given: A box containing 90 discs, numbered from 1 to 90.
To do: The probability that it bears a prime number less than 23, When a disc is drawn from the box.
Solution: Number of discs in the box$=90$
Total possible outcomes$=\{1,\ 2,\ 3,\ 4,.......90\}$
No. of total possible outcomes$=90$
Prime numbers less than 23 are { 2, 3, 5, 7, 11, 13, 17, 19}
Therefore no. of favorable outcomes$=8$
The probability of getting a prime numbered disc $=\frac{No.\ of\ favorable\ outcomes}{Total\ number\ of\ possible\ outcomes}$
$=\frac{8}{90}$
$=\frac{4}{45}$
$\therefore$ Option $( C)$ is correct.
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