Waiting on the platform, a commuter hears an announcement that the train is running five minutes late. He assumes the arrival time may be mo

Question

Waiting on the platform, a commuter hears an announcement that the train is running five minutes late. He assumes the arrival time may be modeled by the random variable T, such that
f(T = t) = {3/5 (5/t)^4 , t ≥ 5
0, otherwise
If given the train arrived in less than 15 minutes, what is the probability it arrived in less than 10 minutes?
А. 62%
B. 73%
C. 88%
D. 91%
E. 96%

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Thành Đạt 5 months 2021-08-21T06:37:29+00:00 1 Answers 8 views 0

Answers ( )

    0
    2021-08-21T06:39:18+00:00

    Answer:

    D. 91%

    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.

    In this question:

    Event A: Less than 15 minutes.

    Event B: Less than 10 minutes.

    We are given the following probability distribution:

    f(T = t) = \frac{3}{5}(\frac{5}{t})^4, t \geq 5

    Simplifying:

    f(T = t) = \frac{3*5^4}{5t^4} = \frac{375}{t^4}

    Probability of arriving in less than 15 minutes:

    Integral of the distribution from 5 to 15. So

    P(A) = \int_{5}^{15} = \frac{375}{t^4}

    Integral of \frac{1}{t^4} = t^{-4} is \frac{t^{-3}}{-3} = -\frac{1}{3t^3}

    Then

    \int \frac{375}{t^4} dt = -\frac{125}{t^3}

    Applying the limits, by the Fundamental Theorem of Calculus:

    At t = 15, f(15) = -\frac{125}{15^3} = -\frac{1}{27}

    At t = 5, f(5) = -\frac{125}{5^3} = -1

    Then

    P(A) = -\frac{1}{27} + 1 = -\frac{1}{27} + \frac{27}{27} = \frac{26}{27}

    Probability of arriving in less than 15 minutes and less than 10 minutes.

    The intersection of these events is less than 10 minutes, so:

    P(B) = \int_{5}^{10} = \frac{375}{t^4}

    We already have the integral, so just apply the limits:

    At t = 10, f(10) = -\frac{125}{10^3} = -\frac{1}{8}

    At t = 5, f(5) = -\frac{125}{5^3} = -1

    Then

    P(A \cap B) = -\frac{1}{8} + 1 = -\frac{1}{8} + \frac{8}{8} = \frac{7}{8}

    If given the train arrived in less than 15 minutes, what is the probability it arrived in less than 10 minutes?

    P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{\frac{7}{8}}{\frac{26}{27}} = 0.9087

    Thus 90.87%, approximately 91%, and the correct answer is given by option D.

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