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Find the probability that a number selected at random from the numbers $30, 31, 32, 33, \text{......}, 60$ is (i) a prime number (ii) a multiple of $6$.
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Solution:(a)
(i) Prime numbers are $31, 37, 41, 43, 47, 53, 59$
$\text{P (a prime number)} = \frac{7}{31}$ (1 Mark)
(ii) Multiples of $6$ are $30, 36, 42, 48, 54, 60$
$\text{P (a multiple of 6)} = \frac{6}{31}$ (1 Mark)
(i) Prime numbers are $31, 37, 41, 43, 47, 53, 59$
$\text{P (a prime number)} = \frac{7}{31}$ (1 Mark)
(ii) Multiples of $6$ are $30, 36, 42, 48, 54, 60$
$\text{P (a multiple of 6)} = \frac{6}{31}$ (1 Mark)