Answer: The required probability is 0.30.
Step-by-step explanation:
Since we have given that
Number of 40-W lightbulbs = 4
Number of 60-W light bulbs = 5
Number of 75-W light bulbs = 6
Total bulbs = 4+5+6 = 15
We need to find the probability that exactly two of the 75-W light bulbs.
so, it becomes,
[tex]\dfrac{^6C_2\times ^9C_1}{^{15}C_3}\\\\=\dfrac{135}{455}\\\\=0.30[/tex]
Hence, the required probability is 0.30.