Answer:
P(A∣D) = 0.667
Step-by-step explanation:
We are given;
P(A) = 3P(B)
P(D|A) = 0.03
P(D|B) = 0.045
Now, we want to find P(A∣D) which is the posterior probability that a computer comes from factory A when given that it is defective.
Using Bayes' Rule and Law of Total Probability, we will get;
P(A∣D) = [P(A) * P(D|A)]/[(P(A) * P(D|A)) + (P(B) * P(D|B))]
Plugging in the relevant values, we have;
P(A∣D) = [3P(B) * 0.03]/[(3P(B) * 0.03) + (P(B) * 0.045)]
P(A∣D) = [P(B)/P(B)] [0.09]/[0.09 + 0.045]
P(B) will cancel out to give;
P(A∣D) = 0.09/0.135
P(A∣D) = 0.667
Answer:
35 homeruns
Step-by-step explanation:
583*0.06=349.8-->34.98-->35
Answer:
x=0.6241
Step-by-step explanation:
Square both sides and end up with
75.24+x=75.8641
isolate x
x=75.8641 - 75.24
x=0.6241
1. Add 5 to both sides
k/3 = 34 + 5
2. Simplify 34 + 5 to 39
k/3 = 39
3. Multiply both side by 3
k = 39 * 3
4. Simplify 39 * 3 to 117
k = 117
10 refers to the number of 100's
multiple 10 by 100 to get 1000
