TWELVE PRACTICE PROBLEMS--PROBABILITY



TWELVE PRACTICE PROBLEMS--PROBABILITY

1. Conditional Probability Problem--Football

From viewing tapes of previous games, a football coach identifies the following six tendencies regarding the next opponent:

a. 70% of the plays are runs to the right (R).

b. 30% of the plays are runs to the left (L).

c. When the play goes to the right, offensive player X has a balanced stance (B) 20% of the time and

d. a shifted stance (S) 80% of the time.

e. When the play goes to the left, offensive player X has a balanced stance 90% of the time and

f. a shifted stance 10% of the time.

Write each item above in probability notation, using the letters R, L, S and B

a. b. c.

d. e. f.

Create a probability table with columns R and L, and rows S and B. Fill in all the numbers.

Compute the following probabilities:

P (L | B) =

P (R | B) =

P (L | S) =

P (R | S) =

Are X's stance and play direction independent? Explain.

You coach the defense: "Always expect a right-side play." What percentage of the time will they be correct?

You coach the defense: "If player X is balanced, expect a left-side play; if player X is shifted, expect a right-side play." What percentage of the time will they be correct?

2. Conditional Probability Problem--Medical Screening Test

Consider a medical screening test to identify cases of an illness that has no symptoms in its early stages. The disease afflicts one-tenth of one per cent of the population. The test is 98% accurate. That is, 98% of people who have the illness will test positive and 2% will test negative (false negatives); and 98% of people who do not have the illness will test negative and 2% will test positive (false positives).

Terminology note: The P (P | D) is called "sensitivity" of the test--its ability to detect when the disease is present. The P (N | W) is called "specificity" of the test--its ability to detect when the disease is not present. These are both 98% in the above example. Sensitivity and specificity need not be equal. Developers of medical tests can manipulate the sensitivity and specificity--when one is increased, the other decreases.

Create a probability table with columns D (disease) and W (well), and rows P (positive) and N (negative). Fill in all the numbers.

Suppose you undergo the screening test and the result is positive. What is the probability that you actually have the disease? What if the test is negative?

3. Conditional Probability Problem--Three Envelopes

There are three envelopes that look alike. One contains two $1 bills, one contains two $100 bills, and one contains one $1 bill and one $100 bill. You are permitted to choose one envelope. Then you are asked to remove one bill from the envelope without looking at the other bill. Suppose that a $100 bill comes out. What is the probability that the other bill is $100? A probability table may be used, but is not necessary.

Here is the same problem in another form. Three index cards are in an envelope. One card is white on both sides. One card is red on both sides. One card is white on one side and red on the other. One card is taken out of the envelope at random and carefully placed on a table so that only one side of the card has been seen. The visible side is red. What is the probability that the other side is also red?

4. Conditional Probability Problem--Three Diseases

Three diseases, A, B and C are mutually exclusive--they cannot occur together. Disease A afflicts one per cent of the population, B afflicts one-half of one per cent of the population, and C afflicts two per cent of the population. Sometimes a disease can be present without the victim having any symptoms. 10% of the A victims have no symptoms, 5% of the B victims have no symptoms, and 25% of the C victims have no symptoms. Symptoms are present only when disease A, B or C is present.

If a person has symptoms, what are the probabilities that he/she has disease A? B? C?

If a person has no symptoms, what are the probabilities that he/she has disease A? B? C?

Use a probability table with columns A, B, C and W (well), and rows S (symptoms) and NS (no symptoms).

5. Probability Problem--Convention Dinner

10,000 guests are attending a convention dinner. 10,000 places are set, with a name card at each place, but the card is face down. The guests are told not to look at the name cards, but to sit anywhere they want. When everyone is seated they are told to look at the name card, and to stand up if they see their own name on the card. What is the probability that at least one person will stand up?

6. Probability Problem--Card-Matching

Two identical shuffled decks of 52 cards are placed face down side by side. The top card of each deck is turned over. If they do not match exactly, the next card of each deck is turned over. If they do not match exactly, the next card of each deck is turned over, etc. What is the probability of an exact match before running out of cards?

7. Probability Problem--The "Monty Hall" or "Let's Make A Deal" Problem

Here is the actual text of the problem as posed to "Ask Marilyn" columnist Marilyn vos Savant in the September 9, 1990 issue of Parade magazine. The question was submitted by Craig F. Whitaker of Columbia, Maryland.

Suppose you're on a game show and you're given the choice of three doors. Behind one door is a car; behind the others, goats. You pick a door, say number 1, and the host, who knows what's behind the doors, opens another door, say number 3, which has a goat. He then says to you, "Do you want to switch to door number 2?" Is it to your advantage to switch your choice?

What is the probability of winning the car if you stick with door number 1?

What is the probability of winning the car if you switch to door number 2?

A probability table may be used--it will have three rows and three columns. Logic can also be used without a probability table.

You may assume that the host will always open a door with a goat in order to prolong the game; that the car is initially placed randomly among the doors by the game-show staff; that the host knows where the car is; and that if the host has a choice as to which door to open, he chooses randomly.

8. Probability Problem--Redundancy

You are looking at a box of old signal flares in a military surplus store. Each one now has an estimated 60% probability of working properly. How many of these must you take on a wilderness backpacking trip in order to be 99% sure of having at least one flare that will work if needed?

9. Probability Problem--Space Shuttle

Suppose that on each space shuttle flight, there is a 3% chance that some member(s) of the crew will not return alive. After how many consecutive successful flights does the probability of all the crew members returning alive drop to 50%?

10. Probability Problem--Rail Crossings

Suppose that the probability of a train passing a road crossing without a fatal accident occurring is 0.999999 ("one in a million" chance of an accident). If four trains pass the crossing each hour of every day, what is the probability of a fatal accident within a ten-year period?

11. Probability Problem--Redundancy

Suppose an aircraft has 100 critical systems, all of which must work if the flight is not to end in disaster. If the reliability of each system is 0.99 (probability of working properly during a given flight) what is the probability of disaster?

Now suppose that each of the 100 critical systems has a backup that will be activated automatically if the primary system fails. What is the probability of a disaster?

12. Probability Problem--Redundancy

Shortly after you begin your new job, you learn that the new person, traditionally, is in charge of organizing the next annual company picnic. You make plans for Saturday, July 20. Because the probability of good picnic weather on any given day during July is only 80%, you are considering a backup date.

What is the probability of a successful picnic if there is no backup date?

What is the probability of a successful picnic if there is one backup date?

Your co-workers tell you that your predecessor was fired for blowing the picnic last year.

What is the probability of a successful picnic if there are two backup dates?

What is the probability of a successful picnic if there are three backup dates?

Comment on the use of July 21 as a backup date.

SOLUTIONS

1. Conditional Probability Problem--Football (see spreadsheet )

a. P(R) = 0.70 b. P(L) = 0.30 c. P(B | R) = 0.20

d. P(S | R) = 0.80 e. P(B | L) = 0.90 f. P(S | L) = 0.10

Probability Table

R L

------ ------

S 0.56 0.03 | 0.59

B 0.14 0.27 | 0.41

------ ------ ------

0.70 0.30 | 1.00

Looking at the completed table does not give a sense of the logical order in which the numbers were entered. Here is the order and rationale:

1.00 lower right corner total is always 1

0.70 given

0.30 given

0.14 given that 20% of the R plays are B

0.56 given that 80% of the R plays are S

0.27 given that 90% of the L plays are B

0.03 given that 10% of the L plays are S

0.59 first-row total

0.41 second-row total

P (L | B) = 0.27 / 0.41 = 0.6585 P (R | B) = 0.14 / 0.41.= 0.3415

Note that this is a pair of conditional complementary probabilities.

P (L | S) = 0.03 / 0/59 = 0.0508 P (R | S) = 0.56 / 0/59 = 0.9492

Another pair of conditional complementary probabilities.

Are X's stance and play direction independent? Explain.

No, they are dependent. R and S are positively related, as are B and L.

P(R | S) > P(R) and P(B | L) > P(B)

R and B are negatively related, as are S and L.

P(R | B) < P(R) and P(S | L) < P(S)

You coach the defense: "Always expect a right-side play." What percentage of the time will they be correct? 70% of the time (it is given that 70% of plays are R).

You coach the defense: "If player X is balanced, expect a left-side play; if player X is shifted, expect a right-side play." What percentage of the time will they be correct?

83% of the time. (Add the probabilities in the two cells that correspond to the positive-dependency combinations.)

2. Conditional Probability Problem--Medical Screening Test

Create a probability table with columns D (disease) and W (well), and rows P (positive test) and N (negative test).

Probability Table

D W

---------- ----------

P 0.00098 0.01998 | 0.02096

N 0.00002 0.97902 | 0.97904

---------- ---------- ----------

0.00100 0.99900 | 1.00000

Looking at the completed table does not give a sense of the logical order in which the numbers were entered. Here is the order and rationale:

1.00000 lower right corner total is always 1

0.00100 given

0.99900 the complement of 0.00100

0.00098 given that 98% of the D people get P test results

0.00002 2% of the D people must get N test results (false negatives)

0.97902 given that 98% of the W people get N test results

0.01998 2% of the W people must get P test results (false positives)

0.02096 P row total

0.97904 N row total

Suppose you undergo the screening test and the result is positive. What is the probability that you actually have the disease? What if the test is negative?

P(D | P) = 0.00098 / 0.02096 = 0.04676

P(D | N) = 0.00002 / 0.97904 = 0.00002043

So even if the test is positive, the probability of actually being sick is still quite low.

3. Conditional Probability Problem--Three Envelopes

This one may be done with a probability table. Symbols:

S1: first bill out of the envelope is a $1 bill

S2: second bill out of the envelope is a $1 bill

C1: first bill out of the envelope is a $100 bill

C2: second bill out of the envelope is a $100 bill

We are looking for P(C2 | C1)

S1 C1

------- -------

S2 0.333 0.167 | 0.500

C2 0.167 0.333 | 0.500

------- ------- -------

0.500 0.500 | 1.000

Looking at the completed table does not give a sense of the logical order in which the numbers were entered. Here is the order and rationale:

1.000 lower right corner total is always 1

0.500 when the first bill is taken out, there is a 50/50 chance between being a $100 bill and a $1 bill, since half of the bills are singles and half are hundreds

0.500 the complement of 0.500

0.333 P(S1 ( S2) is 1/3 since one of the three envelopes has two singles

0.333 P(C1 ( C2) is 1/3 since one of the three envelopes has two hundreds

0.167 by subtraction in the S1 column

0.167 by subtraction in the C1 column

0.500 S2 row total

0.500 C2 row total

What is the probability that the other bill is $100?

P(C2 | C1) = 0.333 / 0.500 = 0.667

Nearly everyone answers this question "50%," but they are wrong. Here's another way of looking at it. When the envelope is selected, there is a 2/3 chance, 0.667, that the two bills in it match. So when the first bill taken out is $100, there is a 0.667 chance that the other will match it and also be $100.

In the three-card case, when the card is selected, there is a 2/3 chance that the card has matching sides. So whatever color is visible, there is a 2/3 chance that the other side is the same.

4. Conditional Probability Problem--Three Diseases (see spreadsheet

If a person has symptoms, what are the probabilities that he/she has disease A? B? C? If a person has no symptoms, what are the probabilities that he/she has disease A? B? C?

A B C W

---------- ---------- ---------- ----------

S 0.00900 0.00475 0.01500 0.00000 | 0.02875

NS 0.00100 0.00025 0.00500 0.96500 | 0.97125

---------- ---------- ---------- ---------- ----------

0.01000 0.00500 0.02000 0.96500 | 1.00000

Looking at the completed table does not give a sense of the logical order in which the numbers were entered. Here is the order and rationale:

1.00000 lower right corner total is always 1

0.01000 given

0.00500 given

0.02000 given

0.96500 by subtraction in the total row--missing number to equal 1

0.00100 given that 10% of the A victims have no symptoms

0.00900 by subtraction in the A column

0.00025 given that 5% of the B victims have no symptoms

0.00475 by subtraction in the B column

0.00500 given that 25% of the C victims have no symptoms

0.01500 by subtraction in the C column

0.00000 given that symptoms are present only when one of the diseases is present--this would make P(S ( W) equal to zero

0.96500 by subtraction in the W column

0.02875 by addition in the S row

0.97125 by addition in the NS row

P(A | S) = 0.00900 / 0.02875 = 0.3130 P(A | NS) = 0.00100 / 0.97125 = 0.001030

P(B | S) = 0.00475 / 0.02875 = 0.1652 P(B | NS) = 0.00025 / 0.97125 = 0.000257

P(C | S) = 0.00150 / 0.02875 = 0.5217 P(C | NS) = 0.00500 / 0.97125 = 0.005148

P(W | S) = 0.00000 / 0.02875 = 0.0000 P(W | NS) = 0.96500 / 0.97125 = 0.993600

Note that the numbers in each column add to 1. Each column is a conditional-probability partition. Consider the implication of the results for people with symptoms. If the diseases have different diagnostic tests, in what order should the diseases be tested for? Consider the implication of the results for people without symptoms. The probabilities of being sick seem reassuringly low for an individual, but what about for a city of a million people?

5. Probability Problem--Convention Dinner

Complementary thinking: P(each person not seeing his/her own name) = 9,999 / 10,000 = 0.9999. By the extended multiplicative rule, the probability of no one seeing his/her own name is 0.9999 to the 10,000 power, or 0.3679. So the probability of at least one person standing up is 0.6321.

6. Probability Problem--Card-Matching

This is the same problem as No. 5. P(each pair not matching) = 51/52 = 0.9808. By the extended multiplicative rule, the probability of no pairs matching is 0.9808 to the 52 power, or 0.3643. So the probability of at least one match is 0.6357.

7. Probability Problem--The "Monty Hall" or "Let's Make A Deal" Problem

Symbols: C1: car behind door 1 H1: host opens door 1

C2: car behind door 2 H2: host opens door 2

C3: car behind door 3 H3: host opens door 3

C1 C2 C3

------- ------- -------

H1 0.000 0.000 0.000 | 0.000

H2 0.167 0.000 0.333 | 0.500

H3 0.167 0.333 0.000 | 0.500

------- ------- ------- -------

0.333 0.333 0.333 | 1.000

Looking at the completed table does not give a sense of the logical order in which the numbers were entered. Here is the order and rationale, assuming that the contestant has chosen door 1 and the host has opened door 3, revealing one of the goats (it does not matter which door you assume the contestant to have chosen, nor the door that the host has opened, as long as there is a goat behind it):

1.000 lower right corner total is always 1

0.333 given that the car is placed randomly by the show staff

0.333 given that the car is placed randomly by the show staff

0.333 given that the car is placed randomly by the show staff

0.000 H1 total--the host will not open door 1 because the contestant chose it

0.000 every entry in row H1 must be zero if the H1 total is zero

0.000 every entry in row H1 must be zero if the H1 total is zero

0.000 every entry in row H1 must be zero if the H1 total is zero

0.000 P(H2 ( C2) is zero--the host will not open the door where the car is

0.000 P(H3 ( C3) is zero--the host will not open the door where the car is

0.333 P(H2 ( C3) subtraction in the C2 column

0.333 P(H3 ( C2) subtraction in the C3 column

0.167 P(H2 ( C1) given that when the host has a choice of which door to open (car behind door 1) he chooses randomly between doors 2 and 3

0.167 P(H3 ( C1) given that when the host has a choice of which door to open (car behind door 1) he chooses randomly between doors 2 and 3

0.500 addition in row H2

0.500 addition in row H3

We are interested in P(C1 | H3), and P(C2 | H3).

P(C1 | H3) = 0.167 / 0.500 = 0.333 P(C2 | H3) = 0.333 / 0.500 = 0.667

You double your chance of winning the car if you switch to door 2!

8. Probability Problem--Redundancy

If you have n flares, by the extended multiplicative rule, P(all your flares failing) is 0.40 to the n power. This is required to be less than 1%. Trial-and-error experimentation can be used:

n 0.4n

---------- ----------

1 0.4

2 0.16

3 0.064

4 0.0256

5 0.01024 Very close, but still a bit over 1%.

6 0.004096 Finally under 1%--six flares are needed.

Algebra solution: 0.4n = 0.01, solve for n.

Solution: n log 0.4 = log 0.01

n = log 0.01 / log 0.4

n = 5.02588

Since flares do not come in fractions (n is a discrete whole-number variable), six flares are needed, with only about 0.41% risk of all failing.

9. Probability Problem--Space Shuttle

In n flights, by the extended multiplicative rule, P(no deaths) is 0.97 to the n power. This is required to be under 50%. Trial-and-error experimentation can be used:

n 0.97n

-------------- --------------

1 0.97

5 0.858734

10 0.737424

15 0.633251

20 0.543794 getting close

22 0.511656 not quite

23 0.496306 finally

Therefore about a 50/50 chance of going 23 missions without a fatality. Note that the ill-fated Challenger was shuttle mission No. 25.

Algebra solution: 0.97n = 0.50, solve for n.

Solution: n log 0.97 = log 0.50

n = log 0.50 / log 0.97

n = 22.7566

Flights do not come in fractions (n is a discrete variable), 23 flights is the answer.

10. Probability Problem--Rail Crossings

Number of crossings in ten years = 4 x 24 x 365.25 x 10 = 350,640.

P(no accidents in ten years) = 0.999999350,640 = 0.704237

P(at least one accident in ten years) = 0.295763 (complement)

11. Probability Problem--Redundancy

Without backup: P(all systems working) = 0.99100 = 0.366032

P(disaster) = 0.633968 (complement)

With backup (redundancy): P(each system failing) = 0.012 = 0.0001

P(each system not failing) = 0.9999 (complement)

P(all systems working) = 0.9999100 = 0.990049

P(disaster) = .009951 (complement)

12. Probability Problem--Redundancy

Without backup date: P(success) = 0.8

With one backup date: P(bad weather both days) = 0.2 x 0.2 = 0.04

P(success) = 0.96

With two backup dates: P(bad weather 3 days) = 0.2 x 0.2 x 0.2 = 0.008

P(success) = 0.9920

With three backup dates: P(bad weather 4 days) = 0.2 x 0.2 x 0.2 x 0.2 = 0.0016

P(success) = 0.9984

July 21 is bad for two reasons. From a statistical viewpoint, it lacks independence with respect to July 20 because weather patterns often persist for more than one day. Also, if it rains on the 20th, the picnic grounds will probably still be wet the next day. Choosing a backup date a week later would be better.

Try doing the "One Backup" case with a probability table.

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