Answer:
a) 0.27 = 27% probability that a randomly chosen card holder has annual income $20,000 or less.
b) 0.778 = 77.8% probability that (s)he carries a balance
Step-by-step explanation:
Conditional Probability
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.
a) What is the probability that a randomly chosen card holder has annual income $20,000 or less?
20% of 30%(carry no balance).
30% of 70%(carry balance). So
![P = 0.2*0.3 + 0.3*0.7 = 0.06 + 0.21 = 0.27](https://tex.z-dn.net/?f=P%20%3D%200.2%2A0.3%20%2B%200.3%2A0.7%20%3D%200.06%20%2B%200.21%20%3D%200.27)
0.27 = 27% probability that a randomly chosen card holder has annual income $20,000 or less.
b) If this card holder has an annual income that is $20,000 or less, what is the probability that (s)he carries a balance?
Conditional probability.
Event A: Annual income of $20,000 or less.
Event B: Carries a balance.
0.27 = 27% probability that a randomly chosen card holder has annual income $20,000 or less
This means that ![P(A) = 0.27](https://tex.z-dn.net/?f=P%28A%29%20%3D%200.27)
Probability of a income of $20,000 or less and balance.
30% of 70%, so:
![P(A \cap B) = 0.3*0.7 = 0.21](https://tex.z-dn.net/?f=P%28A%20%5Ccap%20B%29%20%3D%200.3%2A0.7%20%3D%200.21)
The probability is:
![P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0.21}{0.27} = 0.778](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%20%3D%20%5Cfrac%7B0.21%7D%7B0.27%7D%20%3D%200.778)
0.778 = 77.8% probability that (s)he carries a balance