Probability and Statistics: Solved Problems

Assignment A

1. Electrical Unit Failure Analysis

An electrical unit consists of four components, each subject to failure. We observe this unit at a specified time to determine which components are working and which have failed. What is the sample space associated with this random experiment?

Let: 7TVxAAAAMklEQVQYV2NgIAVwMwoKMQlyMzIASQEg denote working, and 7TVxAAAAOElEQVQYV2NgQAdCrExAIVE2fm4gJcgO denote failed.

The sample space is:

sk1Po1IGBDvwBSXFyLLSZGa0AAAAASUVORK5CYII

r0FVpCrlQq1hbtYj4GRDuBWRUyNYkTTre+ULPpX+

2. Set Operations

Suppose the sample space is: w8MgooHMhmQAAAAASUVORK5CYII= . Let: Ah3KH1ZdVC6IAAAAAElFTkSuQmCC , P7QsOtpxon0ScHGICU69ueneJYv4B4EfzfMfJs+B , and GVqf0R6ASYXx9u0HKKzAAAAABJRU5ErkJggg== .

Determine the sets: rJ4EMhaCNY2UvgAAAABJRU5ErkJggg==

Here: 1D3L2AXFUIWU9OyTQYAAAAASUVORK5CYII=

QAAAABJRU5ErkJggg==

Thus, s1c+AZFgH4yljD1VAAAAAElFTkSuQmCC

So, 5L3W4yP8V+QmWmTt3MOHHfAAAAABJRU5ErkJggg=

3. Natural Numbers Divisibility

Let: Zu4Tlr8b3J6NIswAAAAASUVORK5CYII= , 7TVxAAAAQElEQVQoU2NgoAAIc3Gi6xZgY0cTEuIQ the set of natural numbers divisible by 3, and 7TVxAAAATUlEQVQoU2NgIBuIMDIyMvGjaBfn4WcQ the set of natural numbers divisible by 7TVxAAAAMElEQVQYV2NgwAYEOBgZmbkZBDh5QLIw .

What is the set bx7TwTJXwJa2mL+Rsgppwxf8T9yBSnjBuEKNPqSA ? What is the set hGHjPoBuFDeVx1AAAAAElFTkSuQmCC ? PBR34Bl2PLPcNwRnLAAAAAElFTkSuQmCC and g10FB+vSr06ZwAAAABJRU5ErkJggg==

So, AWxndTgwNzLlAAAAAElFTkSuQmCC and YjQhSzKz6V4PuP9KgV8oUqo88trLhgAAAABJRU5E

4. Newspaper Subscriptions

In a certain residential suburb, e9w0m4wtn2NJ0XQAAAABJRU5ErkJggg== of all households subscribe to the metropolitan newspaper, nUmzt0v+TKNUIbt9UI6UqHVWmfFZvaeth+VnK2oh subscribe to the local paper, and vH6SEAAAAASUVORK5CYII= subscribe to both newspapers. What proportion of households subscribe to exactly one of the two newspapers?

Out of 7TVxAAAAU0lEQVQoU2NgIB8IsQszMIiwMXLASAkB households, 7TVxAAAAOklEQVQYV5XLxwkAMAwDQKX3sv+0sRH5 subscribe to both papers. Thus, xpXd2LCdSmrQrKgAAAABJRU5ErkJggg== subscribe only to the metropolitan paper and 8V93fAwa7nN2iRUAAAAASUVORK5CYII= subscribe only to the local paper. Hence, esDjmIsCnSayu8AAAAASUVORK5CYII= subscribe to exactly one of the two newspapers.

5. Picture Placement Designs

There are nine different locations in which a picture can be added to a text. If four different pictures are to be placed in the text, how many different designs are there?

Since the pictures are different, this is a permutation problem. So, the number of different designs are:

VYSLKZ7sC4kWKk8F5JP745ZS+7pTjH55nuvDBupL

6. Sample Selection

In how many ways can a sample containing 7TVxAAAAN0lEQVQYV2NgwA5EeZmFGRhEOAWA0mI8 non-defective and 7TVxAAAAM0lEQVQYV2NgwAr4mBnZGBiEWPmFWDhB defective parts be chosen from a group containing eRtWcAAAAAElFTkSuQmCC non-defective and 7TVxAAAAKUlEQVQYV2NgQAKHBUUYjjHMYmDgdmBg defective parts?

There are Li1Xva4lQjPX5GPtyACX4DM0w5SzTM5CfKRnVSuv ways of choosing six non-defective parts from the group of twenty-two non-defective parts and CAaWO3F5YWU+rP8eMaSnYI38H8LlZ8l7vEAAAAAS ways of choosing two defective parts from the group of seven defective parts. So, the total number of choices (by the multiplication principle) would be: eQE1eswibwgAAAABJRU5ErkJggg==

7. Barcode Combinations

How many barcodes can be formed using five 7TVxAAAAL0lEQVQYV2NgIAKIczBy8zNIcPJJ8gox ‘s, four 7TVxAAAAO0lEQVQYV2NgQAUCTMIgAUkuMRAlwQ6W ‘s and seven AK5ECtK3AAAAAElFTkSuQmCC ‘s?

Here we use the principle of similar permutations with CTIBfUwAAAABJRU5ErkJggg== and YzJR0aR71szaYNHjMyTHHPAszIsb6fcB03GbHZ6S . Of course, nCm3k83uNTI7EYoBGrhOjm9oLM+PbgIPDyltN30E . So the required number is: PYj8CUiHf6oAhDRkLnTgqbOUTgJZfRbANmdb8Kdv

8. Selection with and without Order

Suppose the number of ways of selecting 7TVxAAAAK0lEQVQYV2NgQALchgEMV85mMjBwFzDw items from a group when order is important is 7TVxAAAAMElEQVQYV2NgIAyEmdh5GLkY+ASZ+QU5 . What would this number be if the order was not important?

+Rj+Y8+SAW+AVryVRiuI0vOAAAAAElFTkSuQmCC Thus, if EMNWvnOPUbW0Owbs5WGcDa3ZkIXVcLjkvqe9Zrxs and r4Wx2QPQIikGHiNl27adhygoTMsypw0+KUG6KVRR , the number would be: 3vCTqADS1jIsnsAAAAAElFTkSuQmCC

Assignment B

1. Dice Roll Probability

Two fair, distinct dice are rolled. What is the probability that the first die comes up 1 given that the sum on the two dice is 5?

Let CEeBgQPDEuYQZ+FphKfkYggPFE2YQZGISYgQQQiP be the event the first die is 1, and let EBtgmziUG2cQHGENpCwCMwQEaAhojBDJHmAhrDwU be the event the second die is 5. We want: wH24puN948rTAAAAAASUVORK5CYII=

2-4. Marital Status Data Analysis

Use this information for problems 2, 3, and 4: The following data on the marital status of vv3sDWZ8LkpZ5gU8AAAAASUVORK5CYII= U.S. adults was found in Current Population Reports:

Single
0VCFxqWRAfAAAAAElFTkSuQmCC
Married
lcOalg9X8MVa1x4MFiCWaVM6nzqLdWu5BdC4NtMV
Widowed
Wm5Edeh09wtpJHVL9W+gQwJQrDk1IoeQAAAABJRU
Divorced
wmIEq0umQAAAABJRU5ErkJggg==
Total
Male
NbX10Vc3rszuQ0gA+FPUIWYtJ5IpD9gko6v6tOtU
1292981340480
Female
7K1pJ6MShbfcxOvIa+BQ1HhWrbWMPl5yxkr8Oxgb
1043055754520
Total23360370941000

2. Probability Calculations

zcCGZb7gDKpxj5Krff9GIQ+lMYBl5w0fAVF7gIQJ ; 7Ao96U9TGuIytAAAAAElFTkSuQmCC

Find AanbJCgU4TwK0IFKKTM5OxfhPFDpKMcZ6QxfM5Nw

According to this sample, 60.3% were married, 48% were male, and 5.2% were divorced females.

3. Conditional Probabilities

(a) AeeA1uZbeHHdAAAAAElFTkSuQmCC

(b) TdlAcK520TuSdijMfSUr7+jmGy5bUpibkJUbnuHH

(c) Probability of being widowed given that the person is a male is 14.6%. The probability that the person is a female given that the person is divorced is 10.4%.

4. Independence of Events

Are 0VCFxqWRAfAAAAAElFTkSuQmCC and NbX10Vc3rszuQ0gA+FPUIWYtJ5IpD9gko6v6tOtU independent events? Justify your answer.

7zemBDu7QpOBUAAAAABJRU5ErkJggg== ; s9waB7MrrS9D7882cyutL0rcH5GUxDIZlfa3hX4h ; Zpqe4eE2AAAAAElFTkSuQmCC

cDz7RJVpyuwAAAAASUVORK5CYII=

Since 2lmgAAAABJRU5ErkJggg== , the events are dependent.

5. Dice Roll Independence

Two fair, distinct dice (one red and one green) are rolled. Let +gJPYnCL9QAAAABJRU5ErkJggg== be the event the red die comes up even and 7TVxAAAAU0lEQVQYV2NgIBcIMgIBEx9IuxQPH4O0 be the event the sum on the two dice is even. Are awI2jdTxnhts2kmVeAiURjmk5pt7bASCSpSGaaws independent events?

By listing the elements in each event we see that: biCuPpEK5n1tkaSjcOyrUWK30lTl2UfvigrGGmwV ; 5KeUX1erEMUS9sbyBCkcw8fgnlSG2sbcrkPUFvFo and DznJxqPU7616jqeNg4Enudbo62B+u8Uyh6CHt6AA . Thus, the events are independent.

6. Congressional Member Probability

For the 107th Congress, 18.7% of the members were senators and 50% of the senators were Democrats. Using the multiplication rule, determine the probability that a randomly selected member of the 107th

Let AU5BA2wahhRzAAAAAElFTkSuQmCC event the member selected is a Democrat and 7TVxAAAAGElEQVQYV2NgoAwECwoKCjmgmYFVEI89 event the member selected is a senator

We want z9+ROZZhlbWriyAQAAAABJRU5ErkJggg== . By the multiplication rule,

6ASRO6LObQe1wdx0DIVZTBAbDhfgNwGd1YfwD9y8

7. Belief in Aliens

According to the Current Population Reports, 52% of U.S. adults are women. Opinion Dynamics Poll published in USA Today shows that 33% of U.S. women and 54% of U.S. men believe in aliens. What percentage of U.S. adults believe in aliens?

Let hXsMhH7n9TS9SERffMopvXvkwKdE+erN6BSQdpdZ event the adult selected believes in aliens and M7+R97xMljBuX6cHwBG3WfgR1G9AcO0ie1HQAAAA event the adult selected is a man. By law of total probability: rOdgAAAABJRU5ErkJggg== ; n5x1XkPJvY8AAAAASUVORK5CYII=

8. Lung Disease and Smoking

According to the American Lung Association, 7% of the population has lung disease. Of the people having lung disease, 90% are smokers. Of the people not having lung disease, 20% are smokers. What are the chances that a smoker has lung disease?

Let 7TVxAAAAU0lEQVQoU2NgGDAgycfIyMSJYb0kL6sY event the person selected at random is a smoker and 7TVxAAAASUlEQVQoU2NgGDDAw8jEjc1yEQ52rG4S event the selected person has lung disease. We want to compute KDGxNrUr1qjePhVcJQciGmmiGWutwqQ9EZLNKrI8 . By Bayes’ Theorem, gcwQmGCdDqnhwAAAABJRU5ErkJggg==