Q. 5.13 You arrive at a bus stop at 10聽... [FREE SOLUTION]

91影视

A First Course in Probability

Sheldon M. Ross

$Math Studyset 91影视 Explanations$ Math

9th Edition 路 432 pages

ISBN: 9780321794772

Chapter 5: Q. 5.13 (page 213)

You arrive at a bus stop at $10$ a.m., knowing that the bus will arrive at some time uniformly distributed between $10$ and $10 : 30$ .
(a) What is the probability that you will have to wait longer than $10$ minutes?
(b) If, at $10 : 15$ , the bus has not yet arrived, what is the probability that you will have to wait at least an additional $10$ minutes?

Short Answer

Expert verified

(a) The probability that the waiting time will be longer than $10$ minutes is $\frac{2}{3}$ .

(b) The probability of waiting at least an additional $10$ minutes is $\frac{1}{3}$ .

Step by step solution

Part (a) Step 1. Given Information.

Here, it is given that the arrival time of bus is uniformly distributed between $10 - 10.30 a.m.$

A passenger arrives at the bus stop at $10 a.m$ .

Part (a) Step 2. Determine the probability density function.

Let $X$ be the random variable that represents the person's waiting time in the bus stop. Then the probability density function is given by:

$f (x) = \frac{1}{b - a} = \frac{1}{30}$

Part (a) Step 3. Calculate the probability that the waiting time of passenger will be more than 10 minutes.

$P (X > 10) = {鈭�}_{10}^{30} f (x) d x = {鈭�}_{10}^{30} \frac{1}{30} d x = \frac{1}{30} {(30)}_{10}^{30} = \frac{1}{30} (30 - 10) = \frac{2}{3}$

Therefore, the probability that the waiting time of passenger will be more than $10$ minutes is $\frac{2}{3}$ .

Part (b) Step 1. Calculate the probability that the waiting time of passenger will be at least additional 10 minutes.

The probability will be:

$P (X 鈮� 25 X > 15) = \frac{P (25 鈮� X 鈮� 30)}{P (X > 15)}$

$= \frac{{鈭�}_{25}^{30} f (x) d x}{{鈭�}_{15}^{30} f (x) d x} = \frac{{鈭�}_{25}^{30} \frac{1}{30} d x}{{鈭�}_{15}^{30} \frac{1}{30} d x} = \frac{{[\frac{1}{30}]}_{25}^{30}}{{[\frac{1}{30}]}_{15}^{30}} = \frac{\frac{1}{30} (30 - 25)}{\frac{1}{30} (30 - 15)} = \frac{1}{3}$

Therefore, the probability that the waiting time of passenger will be at least additional $10$ minutes is $\frac{1}{3}$ .

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91影视

Short Answer

Step by step solution

Part (a) Step 1. Given Information.

Part (a) Step 2. Determine the probability density function.

Part (a) Step 3. Calculate the probability that the waiting time of passenger will be more than 10 minutes.

Part (b) Step 1. Calculate the probability that the waiting time of passenger will be at least additional 10 minutes.

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