Given: A die is tossed three times.

Then the sample space $S={1,2,3,4,5,6}$

Let $\mathrm{P}(\mathrm{A})=$ probability of getting an odd number in first throw.

$\Rightarrow P(A)=3 / 6=1 / 2 .$

Let $\mathrm{P}(\mathrm{B})=$ probability of getting an even number.

$\Rightarrow P(B)=3 / 6=1 / 2$

Now, probability of getting an even number in three times $=1 / 2 \times 1 / 2 \times 1 / 2=1 / 8$

So, the chances of obtaining an odd number at least once

$=1$ – probability of getting an odd number in no throw

$=1$ – probability of getting an even number in three times

$=1-1 / 8$

$\therefore$ Probability of getting an odd number at least once $=7 / 8$