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 tenyear 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 <<PROBTAB1>>)
a. P(R) = 0.70
d. P(S | R) = 0.80
b. P(L) = 0.30
e. P(B | L) = 0.90
Probability Table
R
L
------ -----S
0.56 0.03
|
0.59
B
|
0.41
-----1.00
0.14
-----0.70
0.27
-----0.30
|
c. P(B | R) = 0.20
f. P(S | L) = 0.10
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 positivedependency 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
P
N
D
---------0.00098
W
---------0.01998
0.00002
---------0.00100
0.97902
---------0.99900
|
0.02096
|
0.97904
---------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
0.00100
0.99900
0.00098
0.00002
0.97902
0.01998
0.02096
0.97904
lower right corner total is always 1
given
the complement of 0.00100
given that 98% of the D people get P test results
2% of the D people must get N test results (false negatives)
given that 98% of the W people get N test results
2% of the W people must get P test results (false positives)
P row total
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:
S2:
C1:
C2:
first bill out of the envelope is a $1 bill
second bill out of the envelope is a $1 bill
first bill out of the envelope is a $100 bill
second bill out of the envelope is a $100 bill
We are looking for P(C2 | C1)
S2
C2
S1
------0.333
C1
------0.167 |
0.500
0.167
------0.500
0.333 |
------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 <<PROBTAB2>>
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
---------0.01000
0.00025
---------0.00500
0.00500
---------0.02000
0.96500
---------0.96500
|
|
0.97125
---------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
0.01000
0.00500
0.02000
0.96500
0.00100
0.00900
0.00025
0.00475
0.00500
0.01500
0.00000
0.96500
0.02875
0.97125
lower right corner total is always 1
given
given
given
by subtraction in the total row--missing number to equal 1
given that 10% of the A victims have no symptoms
by subtraction in the A column
given that 5% of the B victims have no symptoms
by subtraction in the B column
given that 25% of the C victims have no symptoms
by subtraction in the C column
given that symptoms are present only when one of the diseases is
present--this would make P(S  W) equal to zero
by subtraction in the W column
by addition in the S row
by addition in the NS row
P(A | S) = 0.00900 / 0.02875 = 0.3130
P(B | S) = 0.00475 / 0.02875 = 0.1652
P(C | S) = 0.00150 / 0.02875 = 0.5217
P(W | S) = 0.00000 / 0.02875 = 0.0000
P(A | NS) = 0.00100 / 0.97125 = 0.001030
P(B | NS) = 0.00025 / 0.97125 = 0.000257
P(C | NS) = 0.00500 / 0.97125 = 0.005148
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:
C2:
C3:
car behind door 1
car behind door 2
car behind door 3
H1:
H2:
H3:
host opens door 1
host opens door 2
host opens door 3
H1
C1
------0.000
C2
------0.000
C3
------0.000
|
0.000
H2
0.167
0.000
0.333
|
0.500
H3
0.167
------0.333
0.333
------0.333
0.000
------0.333
|
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, 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
0.333
0.333
0.333
0.000
0.000
0.000
0.000
0.000
0.000
0.333
0.333
0.167
lower right corner total is always 1
given that the car is placed randomly by the show staff
given that the car is placed randomly by the show staff
given that the car is placed randomly by the show staff
H1 total--the host will not open door 1 because the contestant chose it
every entry in row H1 must be zero if the H1 total is zero
every entry in row H1 must be zero if the H1 total is zero
every entry in row H1 must be zero if the H1 total is zero
P(H2  C2) is zero--the host will not open the door where the car is
P(H3  C3) is zero--the host will not open the door where the car is
P(H2  C3) subtraction in the C2 column
P(H3  C2) subtraction in the C3 column
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:
Solution:
0.4n = 0.01, solve for n.
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
-------------1
5
10
15
20
22
23
0.97n
-------------0.97
0.858734
0.737424
0.633251
0.543794
0.511656
0.496306
getting close
not quite
finally
Therefore about a 50/50 chance of going 23 missions without a fatality. Note that the illfated Challenger was shuttle mission No. 25.
0.97n = 0.50, solve for n.
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.
Algebra solution:
Solution:
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.