I need help ASAP

Question

A pet shop finds that 78% of its customers have a dog and 36% of its customers ordered both a cat and a dog. Find the probability a customer has a cat, given they have a dog.

Answer 1

The probability of a customer having a cat, given they have a dog, is approximately 0.4615 or 45.15%

Variables:

A = customer has a cat

B = customer has a dog

Given:

P(B) = 0.78, customer has a dog

P(A∩B) = 0.36, customer has a cat and a dog

Conditional probability:

P(A | B) = P(A ∩ B) / P(B)

Substitute and simplify:

P(A | B) = 0.36 / 0.78

= 0.4615

Answer 2

Let D = a person has a dog and C = a person has a cat.

We know that P(D) = 0.78 and P(D and C) = 0.36.

P(C l D) = P(D and C) / P(D) = 0.36 / 0.78 = 0.46

Answer 3

P(d) = .78, P(c&d) =.36, find P(c/d)

= 6/7?

use Bayes Theorem: P(a/b)P(b) = P(b/a)P(a) a monk came up with this, so if you're an atheist or not catholic you may want to shut off the screen so your children what's about to come, (come not cum) as math is what the designer used,

so to get down & dirty

P(c/d)P(d) = P(d/c)P(c)

you're on your own Go from here to God Knows Where? make one tiny arithmetic mistake and you're screwed, royally

go with 6/7

answer is between 0 and 1, 6/7 is in there.