Question:
A die is thrown once. The probability of getting an odd number greater than 3 is
(a) $\frac{1}{3}$
(b) $\frac{1}{6}$
(c) $\frac{1}{2}$
(d) 0
Solution:
Total number of outcomes = 6.
Out of the given numbers, odd number greater than 3 is 5.
Numbers of favourable outcomes = 1.
$\therefore \mathrm{P}$ (getting an odd number greater than 3 ) $=\frac{\text { Number of favourable outcomes }}{\text { Number of all possible outcomes }}$
$=\frac{1}{6}$
Thus, the probability of getting an odd number greater than 3 is $\frac{1}{6}$.
Hence, the correct answer is option (b).