Answer :
Answer:
0.087 = 8.7% probability that this person made a day visit.
0.652 = 65.2% probability that this person made a one-night visit.
0.261 = 26.1% probability that this person made a two-night visit.
Step-by-step explanation:
Conditional Probability
We use the conditional probability formula to solve this question. It is
[tex]P(B|A) = \frac{P(A \cap B)}{P(A)}[/tex]
In which
P(B|A) is the probability of event B happening, given that A happened.
[tex]P(A \cap B)[/tex] is the probability of both A and B happening.
P(A) is the probability of A happening.
In this question:
Event A: Made a purchase.
Probability of making a purchase:
10% of 20%(day visit)
30% of 50%(one night)
20% of 30%(two night).
So
[tex]p = 0.1*0.2 + 0.3*0.5 + 0.2*0.3 = 0.23[/tex]
How likely is it that this person made a day visit?
Here event B is a day visit.
10% of 20% is the percentage of purchases and day visit. So
[tex]P(A \cap B) = 0.1*0.2 = 0.02[/tex]
So
[tex]P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0.02}{0.23} = 0.087[/tex]
0.087 = 8.7% probability that this person made a day visit.
A one-night visit?
Event B is a one night visit.
The percentage of both(one night visit and purchase) is 30% of 50%. So
[tex]P(A \cap B) = 0.3*0.5 = 0.15[/tex]
So
[tex]P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0.15}{0.23} = 0.652[/tex]
0.652 = 65.2% probability that this person made a one-night visit.
A two-night visit?
Event B is a two night visit.
The percentage of both(two night visit and purchase) is 20% of 30%. So
[tex]P(A \cap B) = 0.2*0.3 = 0.06[/tex]
Then
[tex]P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0.06}{0.23} = 0.261[/tex]
0.261 = 26.1% probability that this person made a two-night visit.