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Sagot :
Answer:
a-The probability that item A is discovered as a defective item by method 1 is 0.4
b. The probability that item A is discovered as a defective item by method 2 is 0.4
c-As the probability of detecting A as a defective item is same for the two methods, thus it can be stated that each of the methods is equally effective.
Step-by-step explanation:
a-
The first method where the inspector chooses two of the five items randomly, so the total pairs are as given below:
[tex]N_{total}=n(S_{total})\\N_{total}=n(\{AB,AC,AD,AE,BC,BD,BE,CD,CE,DE\})\\[/tex]
As the total experiments in the overall space is 10 so [tex]N_{total}[/tex] is 10.
As A comes in 4 of these experiments, thus the value of [tex]N_{A}[/tex] is 4.
So the Probability of A is given as
[tex]P(A)=\dfrac{N_{A}}{N_{total}}\\P(A)=\dfrac{4}{10}\\P(A)=0.4[/tex]
The probability that item A is discovered as a defective item by method 1 is 0.4.
b-
Now in order to find the probability of finding item A as defective by method 2, first consider the probability of not finding item A as defective.
This is given as
[tex]P(A)=1-P(A)'[/tex]
Here P(A)' is given as
[tex]P(A)'=P(A)'_{1^{st}}\times P(A)'_{2^{nd}}[/tex]
Here
P(A)'_1st is the probability that the first inspector fails to detect A as a defective item which is given as
[tex]P(A)'_{1^{st}}=1-\dfrac{1}{N_{total\ items}}[/tex]
Here N_total items is 5. so the equation becomes
[tex]P(A)'_{1^{st}}=1-\dfrac{1}{N_{total\ items}}\\P(A)'_{1^{st}}=1-\dfrac{1}{5}\\P(A)'_{1^{st}}=\dfrac{5-1}{5}\\P(A)'_{1^{st}}=\dfrac{4}{5}[/tex]
Similarly
P(A)'_2nd is the probability that the second inspector fails to detect A as a defective item which is given as
[tex]P(A)'_{2^{nd}}=1-\dfrac{1}{N_{total\ items}}[/tex]
Here N_total items is 4 as 1 item has already been checked. so the equation becomes
[tex]P(A)'_{2^{nd}}=1-\dfrac{1}{N_{total\ items}}\\P(A)'_{2^{nd}}=1-\dfrac{1}{4}\\P(A)'_{2^{nd}}=\dfrac{4-1}{4}\\P(A)'_{2^{nd}}=\dfrac{3}{4}[/tex]
So the main equation becomes
[tex]P(A)=1-P(A)'\\P(A)=1-(P(A)'_{1^{st}}\times P(A)'_{2^{nd}})[/tex]
Substituting the values give
[tex]P(A)=1-(P(A)'_{1^{st}}\times P(A)'_{2^{nd}})\\P(A)=1-(\dfrac{4}{5}\times \dfrac{3}{4})\\P(A)=1-\dfrac{3}{5}\\P(A)=\dfrac{5-3}{5}\\P(A)=\dfrac{2}{5}\\P(A)=0.4[/tex]
The probability that item A is discovered as a defective item by method 2 is 0.4.
c-
As evident from the above, that the value of probabilities for both the methods is same therefore the overall effectiveness of each of the methods is same.
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