A four - digit number is formed by using the digits 1, 2, 4, 8 and 9 without repitition. If one number is selected from those numbers, then what is the probability that it will be an odd number?
A$\frac{1}{5}$
B$\frac{2}{5}$
C$\frac{3}{5}$
D$\frac{4}{5}$
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B$\frac{2}{5}$
Total number of outcomes = 5 × 4 × 3 × 2 = 120
he number of favourable cases = 2 (4 × 3 × 2) - = 48 (i.e., odd numbers)
herefore,Required probability $=\frac{48}{120}=\frac{2}{5}.$
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