Answer :
Answer:
D. 91%
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.
In this question:
Event A: Less than 15 minutes.
Event B: Less than 10 minutes.
We are given the following probability distribution:
[tex]f(T = t) = \frac{3}{5}(\frac{5}{t})^4, t \geq 5[/tex]
Simplifying:
[tex]f(T = t) = \frac{3*5^4}{5t^4} = \frac{375}{t^4}[/tex]
Probability of arriving in less than 15 minutes:
Integral of the distribution from 5 to 15. So
[tex]P(A) = \int_{5}^{15} = \frac{375}{t^4}[/tex]
Integral of [tex]\frac{1}{t^4} = t^{-4}[/tex] is [tex]\frac{t^{-3}}{-3} = -\frac{1}{3t^3}[/tex]
Then
[tex]\int \frac{375}{t^4} dt = -\frac{125}{t^3}[/tex]
Applying the limits, by the Fundamental Theorem of Calculus:
At [tex]t = 15[/tex], [tex]f(15) = -\frac{125}{15^3} = -\frac{1}{27}[/tex]
At [tex]t = 5[/tex], [tex]f(5) = -\frac{125}{5^3} = -1[/tex]
Then
[tex]P(A) = -\frac{1}{27} + 1 = -\frac{1}{27} + \frac{27}{27} = \frac{26}{27}[/tex]
Probability of arriving in less than 15 minutes and less than 10 minutes.
The intersection of these events is less than 10 minutes, so:
[tex]P(B) = \int_{5}^{10} = \frac{375}{t^4}[/tex]
We already have the integral, so just apply the limits:
At [tex]t = 10[/tex], [tex]f(10) = -\frac{125}{10^3} = -\frac{1}{8}[/tex]
At [tex]t = 5[/tex], [tex]f(5) = -\frac{125}{5^3} = -1[/tex]
Then
[tex]P(A \cap B) = -\frac{1}{8} + 1 = -\frac{1}{8} + \frac{8}{8} = \frac{7}{8}[/tex]
If given the train arrived in less than 15 minutes, what is the probability it arrived in less than 10 minutes?
[tex]P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{\frac{7}{8}}{\frac{26}{27}} = 0.9087[/tex]
Thus 90.87%, approximately 91%, and the correct answer is given by option D.