Solution: We are given that a box contains 24 light bulbs. 2 light bulbs are defective.
Now if a person selects 10 bulbs at random, without replacement.
We are required to find the probability that both defective bulbs will be selected.
Since there are two defective, therefore two defective can be chosen in [tex]\binom{2}{2}[/tex] ways.
Also since 10 bulbs are selected, the remaining 8 bulbs can be selected in [tex]\binom{22}{8}[/tex]
Also, the total number of ways 10 bulbs can be selected is:
[tex]\binom{24}{10}[/tex]
Therefore, the probability that both defective bulbs will be selected is:
[tex]\frac{\binom{2}{2}\binom{22}{8}}{\binom{24}{10}}[/tex]
[tex]\frac{1 \times 319770}{1961256}[/tex]
[tex]0.1630[/tex] rounded to 4 decimal places
Hence, the probability that both defective bulbs will be selected is 0.1630
