Answer:
a) P( | X ) = 0.180
b) P(Y | ) = 0.998
Step-by-step explanation:
Let
P(X) - Probability that he acts hungry
P(Y) - Probability that he had ate dinner,
Given,
P(X | Y) = 0.5
P(X | ) = 0.99
P(Y) = 0.9
a.)
P( | X ) =
= = 0.180
⇒P( | X ) = 0.180
b.)
P(Y | ) =
= = 0.998
⇒P(Y | ) = 0.998
Solution
Note: Area of Kite
From the question we have
Therefore, y = 2
4x -2(3x+7)
Use the distributive property:
4x - 6x -14
Combine like terms:
-2x - 14
Answer:
87.5%
Step-by-step explanation:
Complete question -
75% of the people who eat at Doug's Diner order fish and 90% of the people who order fish also order fries. Given that 80% of all people who eat at Doug's order fries, what is the probability that a randomly chosen customer orders fries without fish?
Solution
Given
People who order fish at the Doug's Diner = 75%
Out of this 75%, 90% also order fries which means that nearly 67.5% ordered fries with the fish.
Given that, 80% of people dining at Doug's Diner order fries
Then percentage or people who ordered only fires and no fish = 80% - 67.5% = 12.5%
The probability of customers who ordered fish = 100 - 12.5% = 87.5%
The relationship is that the 4 in the hundredths place is 10 times greater than the 4 in the thousandths place.