Answer :
Answer:
The probability of event A is 0.2160.
The probability of event B is 0.2153.
Step-by-step explanation:
Assume that the random variable X is defined as the number of defective components in a lot.
It is provided that of the 1060 component 229 are defective.
The probability of selecting a defective component is:
[tex]P(X)=\frac{229}{1060}=0.2160[/tex]
The proportion of defective components in a lot of 1060 is 0.2160.
It is provided that two components are selected to be tested.
Assuming the selection were without replacement.
A = the first component drawn is defective
B = the second component drawn is defective
- Compute the probability of event A:
The probability of selecting a defective component from the entire lot
of 1060 component is 0.2160.
Thus, the probability of event A is 0.2160.
- Compute the probability of event B:
According to event A, the first component selected was defective.
So now there are 228 defective components among 1059
components.
[tex]P(B)=\frac{228}{1059}= 0.2153[/tex]
Thus, the probability of event B is 0.2153.
Both the probabilities are almost same.
This implies that the probability of selecting a defective component from the entire population of these components is approximately 0.2160.
Answer:
Required Probability = 0.0467
Step-by-step explanation:
We are given that a lot of 1060 components contains 229 that are defective.
Two components are drawn at random and tested.
A = event that the first component drawn is defective
B = event that the second component drawn is defective
So,P(first component drawn is defective,A)=(No. of defective component)÷
(Total components)
P(A) = 229/1060
Similarly, P(second component drawn is defective,A) = 229/1060
Therefore, P(both component drawn defective) = [tex]\frac{229}{1060} * \frac{229}{1060}[/tex] = 0.0467 .