Answer:
a. P(X = 0) = 0.134
P(X = 1) = 0.271
P(X = 2) = 0.272
P(X = 3) = 0.1812
b. P(X > 3) = 0.142
Step-by-step explanation:
The given parameters are
Number of defective component per 1000 = 8
Therefore, the probability that the specific component is defective = 8/1000 = 0.008
The binomial probability is given as follows;
[tex]P(X = r) = \dbinom{n}{r}p^{r}\left (1-p \right )^{n-r}[/tex]
[tex]P(X = 0) = \dbinom{250}{0} \left (\dfrac{1}{125} \right )^{0}\left (\dfrac{124}{125} \right )^{250-0} = 0.134[/tex]
[tex]P(X = 1) = \dbinom{250}{1} \left (\dfrac{1}{125} \right )^{1}\left (\dfrac{124}{125} \right )^{249} = 0.271[/tex]
[tex]P(X = 2) = \dbinom{250}{2} \left (\dfrac{1}{125} \right )^{2}\left (\dfrac{124}{125} \right )^{248} = 0.272[/tex]
[tex]P(X = 3) = \dbinom{250}{3} \left (\dfrac{1}{125} \right )^{3}\left (\dfrac{124}{125} \right )^{247} = 0.1812[/tex]
The probability that the box contains at least 3 defectives is the probability P(X > 3) = 1 - (P(X = 3) + P(X = 2) + P(X = 1) + P(X = 0))
Which gives;
P(X > 3) = 1 - (0.1812 - 0.271 - 0.272 - 0.134) = 0.142
