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Sagot :
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
[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.
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
[tex]P = 0.2*0.3 + 0.3*0.7 = 0.06 + 0.21 = 0.27[/tex]
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 [tex]P(A) = 0.27[/tex]
Probability of a income of $20,000 or less and balance.
30% of 70%, so:
[tex]P(A \cap B) = 0.3*0.7 = 0.21[/tex]
The probability is:
[tex]P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0.21}{0.27} = 0.778[/tex]
0.778 = 77.8% probability that (s)he carries a balance
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