A box contains 19 balls bearing numbers 1, 2, 3, ..., 19 respectively.

Question:

A box contains 19 balls bearing numbers 1, 2, 3, ..., 19 respectively. A ball is drawn at random from the box. Find the probability that the number on the ball is

(i) a prime number

(ii) an even number

(iii) a number divisible by 3.

Solution:

Total number of possible outcomes $=19$

(i) The prime numbers between 1 and 19 are 2, 3, 5, 7, 11, 13, 17 and 19 .

Total number of primes $=8$

$\therefore \mathrm{P}_{(\text {prime number })}=\frac{8}{19}$

(ii) The even numbers between 1 and 19 are $2,4,6,8,10,12,14,16$ and $18 .$

Total number of even numbers $=6$

$\therefore \mathrm{P}_{(\text {even number })}=\frac{9}{19}$

(iii) The numbers between 1 and 19 which are divisible by 3 are $3,6,9,12,15$ and $18 .$

Total number of possible outcomes $=6$

$\therefore \mathrm{P}_{(\text {number }}$ divisble by 3$)=\frac{6}{19}$

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