The numbers divisible by 3 are multiples of 3.
Answer:
31.82% probability that this day would be a winter day
Step-by-step explanation:
We use the conditional probability formula to solve this question. It is
![P(B|A) = \frac{P(A \cap B)}{P(A)}](https://tex.z-dn.net/?f=P%28B%7CA%29%20%3D%20%5Cfrac%7BP%28A%20%5Ccap%20B%29%7D%7BP%28A%29%7D)
In which
P(B|A) is the probability of event B happening, given that A happened.
is the probability of both A and B happening.
P(A) is the probability of A happening
In this question:
Event A: Rain
Event B: Winter day
Probability of rain:
0.42 of 0.25(winter), 0.23 of 0.25(spring), 0.16 of 0.25(summer) or 0.51 of 0.25(fall).
So
![P(A) = 0.42*0.25 + 0.23*0.25 + 0.16*0.25 + 0.51*0.25 = 0.33](https://tex.z-dn.net/?f=P%28A%29%20%3D%200.42%2A0.25%20%2B%200.23%2A0.25%20%2B%200.16%2A0.25%20%2B%200.51%2A0.25%20%3D%200.33)
Intersection:
Rain on a winter day, which is 0.42 of 0.25. So
![P(A \cap B) = 0.42*0.25 = 0.105](https://tex.z-dn.net/?f=P%28A%20%5Ccap%20B%29%20%3D%200.42%2A0.25%20%3D%200.105)
If you were told that on a particular day it was raining in Vancouver, what would be the probability that this day would be a winter day?
![P(B|A) = \frac{0.105}{0.33} = 0.3182](https://tex.z-dn.net/?f=P%28B%7CA%29%20%3D%20%5Cfrac%7B0.105%7D%7B0.33%7D%20%3D%200.3182)
31.82% probability that this day would be a winter day
Answer:
B
Step-by-step explanation:
P (A and B) = P(A) * P(B)...answer is D