# Croq Pain

Only available on StudyMode
• Topic: Probability theory, Cumulative distribution function, Discrete probability distribution
• Pages : 10 (1672 words )
• Published : February 15, 2013

Text Preview
BUS 260
Homework 2 Solutions
Due before the start of class, Wednesday January 16.

1. In Cook County, each day is either sunny or cloudy. If a day is sunny, the following day will be sunny with probability 0.60. If a day is cloudy, the following day will be cloudy with probability 0.70. Suppose it is cloudy on Monday.

a) What is the probability that it will be sunny on Wednesday? There are two mutually exclusive ways that it could end up being sunny on Wednesday. P(Sunny Wednesday) = P(Sunny Tuesday AND Sunny Wednesday) + P(Cloudy Tuesday AND Sunny Wednesday)

Since the weather on consecutive days is related, we use the 4th law: P(Sunny Tuesday AND Sunny Wednesday) = P(Sunny Tuesday) * P(Sunny Wednesday | Sunny Tuesday)
= .30 * .60 = .18
Similarly,
P(Cloudy Tuesday and Sunny Wednesday) = P(Cloudy Tuesday) * P(Sunny Wednesday | Cloudy Tuesday)
= 0.70 * 0.30 = 0.21
So, P(Sunny Wednesday) = .18 + .21 = .39. There’s a 39% chance that it will be sunny on Wednesday. b) What is the probability that it will be sunny on both Tuesday and Wednesday? We calculated this while solving part (a). The probability that both days will be sunny is 0.18. c) It turns out to be sunny on Wednesday. What is the probability that it was also sunny on Tuesday? Since we know it was sunny on Wednesday, we want to calculate the conditional probability: P(Sunny on Tuesday | Sunny on Wednesday)

Using the 4th law,
= P(Sunny on Tuesday AND Sunny on Wednesday) / P(Sunny on Wednesday) = .18/39 = .46. There’s a 46% chance that it was sunny on Tuesday.

2. A job fair was held at the Student Union. 25% of the students who attended received job offers. Of all of the students at the job fair, 40% were from the College of Business. Among these business students, 50% received job offers.

Let J be the event that a student is offered a job. Let B be the event that the student is from the College of Business.
a)

Are events J and B independent? Why or why not?

Events J and B are not independent. If B occurs it is more likely that J occurs. b)

Are events J and B mutually exclusive? Why or why not?

Events J and B are not mutually exclusive. It’s possible for both to occur (that is, a Business student receives a job offer).
c)

Joe, who is not a business student, attended the job fair. What is the probability that he received a job offer?

We want to find P(J | Not B).
Using the 4th law,
= P(J AND Not B) / P(Not B)
The denominator is simply .6, since we know 40% of the students are business students. The numerator takes a bit more work. If we apply the 4th law to the numerator, we have: P(J AND Not B) = P(Not B) × P(J | Not B) which leads us in a circle, since P(J | Not B) is exactly what we are trying to find in the first place!

Instead, we need to back into using the fact that
P(J) = P(J AND B) + P(J AND Not B)
So, if we re-arrange this,
P( J AND Not B) = P(J) – P(J AND B)
= P(J) – P(B) × P(J | B)
= .25 – .4 × .5 = .25 – .20 = .05
Thus,
P(J | Not B) = .05/.6 = .083.

(The two mutually exclusive ways of getting a job offer)

d)

Another student, Samantha, received a job offer. What is the probability that she is a Business student?

P(B | J) = P(B and J)/P(J)
We saw from part (c) that P(B and J) = .2. So,
P(B | J) = P(B and J)/P(J) = .2/.25 = .8
If Samantha received a job offer, there’s an 80% chance that she is a Business student.

3. Consider the random variable T which has the following distribution: Value Probability
3
.5
6
.1
7
.25
10
.1
15
.05

a) Graph the distribution of T.
Probability
0.6

0.5

Probability

0.4

0.3

0.2

0.1

0
0

1

2

3

4

5

6

7

8

9

10

11

12

13

14

15

Value

b) Compute P(T ≤ 7).
P(T ≤ 7) = P(T = 3) + P(T = 6) + P(T = 7) = 0.5 + 0.1 + 0.25 = .85 c) Compute P(3 ≤ T ≤ 10).
P(3 ≤ T ≤ 10) = P(T = 3) + P(T = 6) + P(T = 7) + P(T = 10) = 0.5 + 0.1 + 0.25 + .1 = .95 d) Compute P(T ≥ 10 | T ≥ 6).
P(T ≥ 10 | T ≥ 6) = P(T...