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:
total area = 99m² + 39.4m²
= 138.4m²
Answer:
2 x54 + 2 x 54
108 + 108
216
Step-by-step explanation:
At the end of the zeroth year, the population is 200.
At the end of the first year, the population is 200(0.96)¹
At the end of the second year, the population is 200(0.96)²
We can generalise this to become at the end of the nth year as 200(0.96)ⁿ
Now, we need to know when the population will be less than 170.
So, 170 ≤ 200(0.96)ⁿ
170/200 ≤ 0.96ⁿ
17/20 ≤ 0.96ⁿ
Let 17/20 = 0.96ⁿ, first.
log_0.96(17/2) = n
n = ln(17/20)/ln(0.96)
n will be the 4th year, as after the third year, the population reaches ≈176