A die is tossed four times. The probability of getting an odd number at least once, is

\(\frac{15}{16}\)

Probability of getting an odd number in a single throw of a die = $\frac{3}{6}​=\frac{1}{2​}$

Similarly, probability of getting an even number = $\frac{3}{6}​=\frac{1}{2​}$

as there are 6 outcomes → 1, 2, 3, 4, 5, 6

Probability of getting an even number four times = $\frac{1}{2​}×\frac{1}{2​}×\frac{1}{2​}×\frac{1}{2​}=\frac{1}{16}$

∴ Probability of getting an odd number at least once

=1− Probability of getting an even number four times

$1-\frac{1}{16}=\frac{16-1}{16}=\frac{15}{16}$

Option 4 is correct.