Math 115 A - Section 3 - Test 1



Math 115 A - Section 3 - Test 1 Name:___________________

04 October 2000

In order to get full credit, you must show all of the steps and all of the work that you use to solve each problem.

A list of formulas that you may find useful is provided on the last page of this test.

|Question |Points |Score |

|1 |30 | |

|2 |7 | |

|3 |15 | |

|4 |12 | |

|5 |18 | |

|6 |18 | |

|Total |100 | |

1. You are applying for jobs and have been interviewed by only two different companies, called Alpha Co and Beta Co respectively. You estimate that there is a 60% chance that you will get an offer from Alpha Co and a 45% chance that you will get an offer from Beta Co. You also know that there is a non-zero probability that you will get no offer at all.

Note: the questions (i) to (v) below are independent. If you do not know how to answer one of them, move on to the next one.

i) (4 points) Set up a sample space for this problem and use the information given above to draw a diagram showing the different events (that you will have to define).

A: event that you get a job from Alpha Co. B: event that you get a job from Beta Co.

N: event that you get no job offer.

S = {A, B, N}

ii) Assume that there is a 25% chance that you will get an offer from both companies.

• (4 points) What is the probability that you will get at least one offer ? Explain.

We know that P(A) = 0.6, P(B) = 0.45, and P(A(B) = 0.25. We want

P(A(B) = P(A) + P(B) - P(A(B) = 0.6 + 0.45 – 0.25 = 0.80.

There is an 80% chance that you get at least one offer.

• (4 points) What is the probability that you will get no offer at all ? Explain.

We want P[ (A(B)C ]=1- P(A(B) = 1-0.8 = 0.2.

There is a 20% chance that you get no offer at all.

iii) (5 points) Assume that there is a 10% chance that you will get no offer at all. What is the probability that you get both offers ? Justify your answer.

We know P[ (A(B)C ] = 0.1. Thus, P(A(B) = 1 - P[ (A(B)C ] = 1 – 0.1 = 0.9.

We want P(A(B) = P(A) + P(B) - P(A(B) = 0.6 + 0.45 – 0.9 = 0.15.

There is a 15% chance that you will get both offers.

iv) Assume that the events of getting an offer from Alpha Co and getting an offer from Beta Co are independent.

• (4 points) What is the probability that you will get both offers ? Explain.

Since A and B are independent, P(A(B) = P(A) ( P(B) = 0.6 ) ( 0.45 = 0.27.

There is a 27% chance that you will get both offers.

• (3 points) Explain in real-world terms whether you believe that getting an offer from Alpha Co and getting an offer from Beta Co are independent.

These events should not be independent if we assume that getting an offer is related to individual performance. One can indeed consider that a very marketable person will get offers if she satisfies a series of predetermined criteria. It is likely that different companies use similar criteria. The result of one interview should then be related to the result of another one.

v) (6 points) Assume that there is a 20% chance that you will get both offers. If you know that you are going to get an offer from Alpha Co, what is the probability that you will also get an offer from Beta Co ?

We know P(A(B) = 0.2. We want P(B|A) .

[pic]

There is a 33.33% chance that you will get an offer from Beta Co., given that you got an offer from Alpha Co.

2. Consider two events R and W.

i) (3 points) Shade the event W C on the diagram below.

ii) (4 points) Give a formula for the event shaded below. Also, describe in words what this event corresponds to.

This event is R(WC.

It corresponds to the event

That “R occurs and W does

not occur”.

3. Assume that the random variable X can only take values 0, 1, 2 and 3, with probabilities: P(X=0) = 0.15, P(X=1) = 0.35, and P(X=2) = 0.10.

i) (10 points) Find the expected value of X. Explain what this number means.

[pic]

This means that if we were to pick values of X according to the above probabilities, X would, on average, be equal to 1.75.

ii) (5 points) Find P(X < 4) and P(0 ( X ( 1).

Since X always takes values which are less than 4, P(X < 4) = 1.

P(0 ( X ( 1) = P(X=0) + P(X=1) = 0.15 + 0.35 = 0.5

4. (12 points) You are a loan officer and have to decide whether you should foreclose on one of your clients. You are given the following information: the full value of this person's loan is $700,000; the foreclosure value is $300,000 and the default value is $100,000. You looked at past bank records and you estimated that the probability of success of an attempted workout is 0.45. Use this information to decide whether you recommend workout or foreclosure for this loan. Explain your decision.

Let X be the amount of money that the bank you work for can expect to receive after a workout arrangement. We compute E(X), the expected value of X, and compare it to the foreclosure value. Let S be the event that a loan is successfully paid back after a workout arrangement, and let F be the event that it is not.

[pic]

Since the expected value of X is greater than the foreclosure value, we recommend to work out an arrangement for this loan.

5. You have a coin, which is such that the probability of getting a head is 0.55.

i) (3 points) Is this a fair coin ? Explain.

This cannot be a fair coin. If it were, the probability of getting a head would be 0.5.

ii) (15 points) Let X be the number of heads you obtain when you toss the coin three times. You can assume that the tosses are independent from one another. Find P(X=1). Hint: set up a sample space for the experiment of tossing the coin three times and identify the events for which X=1. Then, find the corresponding probabilities.

Let H be the event that we get a head on a toss, and let T be the event that we get a tail. The sample space for the experiment of tossing the coin three times is

S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}. Only three of these events, namely HTT, THT and TTH correspond to X = 1.

Since tosses are independent from each other,

P(HTT ) = P(H)(P(T)(P(T) = 0.55 ( 0.45 ( 0.45 ( 0.111

Since the events HTT, THT and TTH are mutually exclusive,

P(X=1) = P(HTT ( THT ( TTH ) = P(HTT ) + P(THT ) + P(TTH )

= 3 ( P(HTT ) ( 0.334

6. You are organizing a conference and know that all of your out-of-town participants will come by plane. There are three airlines they can use, called Airline One, Airline Two and Airline Three. Let A be the event that a randomly selected out-of-town participant flies on Airline One, let B be the event that she flies on Airline Two and C be the event that she flies on Airline Three. You know that P(A) = 0.45, P(B) = 0.30, and P(C) = 0.25. You also know that a flight arriving to your city is on time with the following probabilities: 80% for Airline One, 75% for Airline Two and 60% for Airline Three.

i) (4 points) Let S be the event that a randomly selected out-of-town participant arrives on time. Describe in words the meaning of P(S | A).

P(S | A) is the probability that a randomly selected participant arrives on time, given that she flew with Airline One.

ii) (14 points) Find the probability that a randomly selected out-of-town participant arrives on time. Justify your answer.

We want P(S). Since the events A, B and C make a partition of the sample space, we have

[pic]

There is a 73.5% chance that a randomly selected participant will arrive on time.

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