Thirty percent of credit card holders carry no monthly balance, while 70% do. Of those card holders carrying a balance, 30% have annual inco

Question

Thirty percent of credit card holders carry no monthly balance, while 70% do. Of those card holders carrying a balance, 30% have annual income $20,000 or less, 40% between $20,001 – $50,000, and 30% over $50,000. Of those card holders carrying no balance, 20%, 30%, and 50% have annual incomes in these three respective categories.
a) What is the 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?

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Thu Thảo 5 years 2021-08-26T03:37:40+00:00 1 Answers 31 views 0

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  1. MichaelMet
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    2021-08-26T03:39:14+00:00
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    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)}

    In which

    P(B|A) is the probability of event B happening, given that A happened.

    P(A \cap B) 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

    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

    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

    The probability is:

    P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0.21}{0.27} = 0.778

    0.778 = 77.8% probability that (s)he carries a balance

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