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Total no. of bulbs = 24

Defective bulbs = 4

Out of 24 light bulbs, 4 are defective.

Let the first person selecting 10 bulbs be A and the second person taking the remaining 14 bulbs be B.

The random selection of10 bulbs from the box of 24 bulbs can be done in\(^{24}{C_{10}}\)ways.

Therefore, the probability that person A will select all 4 defective bulbs is,

\(\begin{aligned}{l}\frac{{^4{C_4}{ \times ^{20}}{C_6}}}{{^{24}{C_{10}}}}\\ = 0.01976\end{aligned}\)

Similarly, the probability that person B will select all 4 defective bulbs is,

\(\begin{aligned}{l}\frac{{^4{C_0}{ \times ^{20}}{C_{10}}}}{{^{24}{C_{10}}}}\\ = 0.09420\end{aligned}\)

As there are two independent occurrences, we will add these two probabilities.

\(\begin{aligned}{c}{{\rm{P}}_{\rm{r}}} &= 0.01976 + 0.09420\\ &= 0.1140\end{aligned}\)

Thus, the required probability is 0.1140.