Respuesta :
Answer:
(a) Probability that a randomly chosen item is defective and cannot be repaired is 8%.
(b) Probability that exactly 2 of 20 randomly chosen items are defective and cannot be repaired is 0.2711.
Step-by-step explanation:
We are given that of the items manufactured by a certain process, 20% are defective. Of the defective items, 60% can be repaired.
Let Probability that item are defective = P(D) = 0.20
Also, R = event of item being repaired
Probability of items being repaired from the given defective items = P(R/D) = 0.60
So, Probability of items not being repaired from the given defective items = P(R'/D) = 1 - P(R/D) = 1 - 0.60 = 0.40
(a) Probability that a randomly chosen item is defective and cannot be repaired = Probability of items being defective [tex]\times[/tex] Probability of items not being repaired from the given defective items
= 0.20 [tex]\times[/tex] 0.40 = 0.08 or 8%
So, probability that a randomly chosen item is defective and cannot be repaired is 8%.
(b) Now we have to find the probability that exactly 2 of 20 randomly chosen items are defective and cannot be repaired.
The above situation can be represented through Binomial distribution;
[tex]P(X=r) = \binom{n}{r}p^{r} (1-p)^{n-r} ; x = 0,1,2,3,.....[/tex]
where, n = number of trials (samples) taken = 20 items
r = number of success = exactly 2
p = probability of success which in our question is % of randomly
chosen item to be defective and cannot be repaired, i.e; 8%
LET X = Number of items that are defective and cannot be repaired
So, it means X ~ [tex]Binom(n=20, p=0.08)[/tex]
Now, Probability that exactly 2 of 20 randomly chosen items are defective and cannot be repaired is given by = P(X = 2)
P(X = 2) = [tex]\binom{20}{2} \times 0.08^{2} \times (1-0.08)^{20-2}[/tex]
= [tex]190 \times 0.08^{2} \times 0.92^{18}[/tex]
= 0.2711
Therefore, probability that exactly 2 of 20 randomly chosen items are defective and cannot be repaired is 0.2711.
