Answer: 0.2501
Step-by-step explanation:
Given : The proportion of the bulbs in the lot are defective = 20%= 0.20
Sample size : n =15
Let x be the binomial variable that represents the defective bulbs.
Binomial probability formula : [tex]^nC_xp^x(1-p)^x[/tex]
The probability that exactly 3 bulbs from the sample are defective : P(X=3)
[tex]=^{15}C_3(0.20)^{3}(1-0.20)^{12}\\\\=\dfrac{15!}{3!12!}(0.20)^3(0.80)^{12}=0.2501[/tex]
Required probability = 0.2501
