Question

Consider randomly selecting a student at a certain university, and let A denote the event that the selected individual has a Visa credit card and B be the analogous event for a MasterCard where P(A) = 0.45, P(B) = 0.35, and P(A ❩ B) = 0.30. Calculate and interpret each of the following probabilities (a Venn diagram might help). (Round your answers to four decimal places.)(a) P(B | A)(b) P(B' | A)(c) P(A | B)(d) P(A' | B)

Answer 1

Answer:

a) $P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0.3}{0.45}= 0.667$

Represent the probability that the event B occurs given that the event A occurs first

b) $P(B'|A) = \frac{0.15}{0.45}=0.333$

Represent the probability that the event B no occurs given that the event A occurs first

c) $P(A|B) = \frac{P(A \cap B)}{P(B)} = \frac{0.3}{0.35}= 0.857$

Represent the probability that the event A occurs given that the event B occurs first

d) $P(A'|B) = \frac{0.05}{0.35}=0.143$

Represent the probability that the event A no occurs given that the event B occurs first

Step-by-step explanation:

For this case we have the following probabilities given for the events defined A and B

$P(A) = 0.45, P(B) = 0.35, P(A \cap B) =0.30$

For this case we can begin finding the probability for the complements:

$P(B') =1-P(B) = 1-0.35= 0.65$

$P(A') =1-P(A) = 1-0.45= 0.55$

For this case we are interested on the following probabilities:

Part a

$P(B|A)$

For this case we can use the Bayes theorem and we can find this probability like this:

$P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0.3}{0.45}= 0.667$

Represent the probability that the event B occurs given that the event A occurs first

Part b

$P(B'|A) = \frac{P(B' \cap A)}{P(A}$

And for this case we can find $P(B' \cap A) =P(A) -P(A\cap B)= 0.45-0.3=0.15$

And if we replace we got:

$P(B'|A) = \frac{0.15}{0.45}=0.333$

Represent the probability that the event B no occurs given that the event A occurs first

Part c

$P(A|B) = \frac{P(A \cap B)}{P(B)} = \frac{0.3}{0.35}= 0.857$

Represent the probability that the event A occurs given that the event B occurs first

Part d

$P(A'|B) = \frac{P(A' \cap B)}{P(B}$

And for this case we can find $P(A' \cap B) =P(B) -P(A\cap B)= 0.35-0.3=0.05$

And if we replace we got:

$P(A'|B) = \frac{0.05}{0.35}=0.143$

Represent the probability that the event A no occurs given that the event B occurs first