Answer:
The probability is 0.0008.
Step-by-step explanation:
Let X represents the event of defective bulb,
Given, the probability of defective bulb, p = 20 % = 0.2,
So, the probability that bulb is not defective, q = 1 - p = 0.8,
The number of bulbs drawn, n = 10,
Since, binomial distribution formula,
[tex]P(x=r) = ^nC_r p^r q^{n-r}[/tex]
Where, [tex]^nC_r = \frac{n!}{r!(n-r)!}[/tex]
Hence, the probability that exactly 7 bulbs from the sample are defective is,
[tex]P(X=7)=^{10}C_7 (0.2)^7 (0.8)^{10-7}[/tex]
[tex]=120 (0.2)^7 (0.8)^3[/tex]
[tex]=0.000786432[/tex]
[tex]\approx 0.0008[/tex]
