A distillery is planning a production run of 600 litere of a blended whiskey Three components burbons are mixed to form the final product. Burbons i iz and 3 costs are Rs 10, RS is and Rs 20 per liter respectively. The blending reape calls for burbon 1 to be 24 time the amount of burbon 3 in the final blend. The total cost of the componends blend should equal Rs 8250. Determine the quantity of each burbon whicb_sbowd be included in the final blend using determinant method.
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Answer:
Let us consider the probabilities of production
P(A)=
3500
500
=
7
1
P(B)=
3500
1000
=
7
2
P(C)=
3500
2000
=
7
4
And,
Probability of defective pipes
By plant A=P(A/E)=0.005
By plant B=P(B/E)=0.008
By plant C=P(C/E)=0.010
Now,
from Bayes theorem
Probability of defective pipe from first plant
P(E/A)=
[P(A).P(A/E)+P(B).P(B/E)+P(C).P(C/E)]
[P(A)×P(A/E)]
=
[
7
1
×0.005+
7
2
×0.008+
7
3
×0.010]
[
7
1
×0.005]
=
[
7
0.005
+
7
2(0.008)
+
7
3(0.010)
]
[
7
0.005
]
=
0.005+0.016+0.040
0.005
=
0.061
0.005
=
61
5
So, the probability is from First factory is
61
5
mark as brainliest please
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