Department of Industrial Engineering
IE 202: Engineering Statistics
Example Questions
Spring 2012
1. Twenty workers are to be assigned to 20 different jobs, one to each job. How many different assignments
are possible?
Solution:
The first worker chooses 1 from 20 jobs. There are 19 jobs and 19 workers left. The second worker
chooses 1 from 19 jobs. There are 18 jobs and 18 workers left, and so on. So the answer is 20!.
2. John, Jim, Jay and Jack have formed a band consisting of 4 instruments. If each of the boys can play
all 4 instruments, how many different arrangements are possible? What if John and Jim can play all 4
instruments but Jay and Jack can each play only piano and drums?
Solution
If each of the boys can play all 4 instruments, then there are 4! ways for them to play together.
If Jay and Jack can only play piano and drum, then John and Jim must play the other 2 instruments.
Then there are 2! × 2! = 4 ways for them to play.
3. In how many ways can 8 people be seated in a row if
(a) there are no restrictions on the seating arrangement?
(b) persons A and B must sit next to each other?
(c) there are 4 men and 4 women and no 2 men or 2 women can sit next to each other?
(d) there are 5 men and they must sit next to each other?
(e) there are 4 married couples and each couple must sit together?
Solution
(a) 8!
(b) First consider A and B as 1 person because they must sit next to each other, then there are 7!
ways for them to sit. Next consider that however they sit, A can sit to the left of B or B can sit
to the left of A. So the answer is 2 × 7!.
(c) Let M represent ’Man’ and W represent ’Woman’. Here the arrangement must be M-W-M-W-MW-M-W OR W-M-W-M-W-M-W-M. In each case there are 4 possible places for the 4 women to
sit and 4 possible places for the 4 men to sit. Therefore the answer is: 2 × 4! × 4!.
(d) First consider the 5 men as 1 person because they must sit next to each other. there are 8 people,
so three of them are woment. Therefore there are 4! such seating arrangements. Now consider that
no matter where the women are seated, there are 5! ways that the five men can sit next to each
other. The answer is 4! × 5!.
(e) Consider every married couple as 1 person, there are 4! ways for the couple to sit. For every couple,
the wife can sit to the left of the husand or the husand can sit to the left of the wife, so there are
2 ways for them to sit together. The answer is 4! × 24 .
1
4. Seven different gifts are to be distributed among 10 children. How many distinct results are possible if
no child is to receive more than one gift?
Solution
If the gifts are unique. The first gift can be given to one of 10 children. The second gift goes to one of
the remaining 9 children, and so on. The answer is
10 × 9 × 8 × 7 × 6 × 5 × 4 =
10!
3!
If the gifts are the same. The same logic applies, but since the gifts are the same, some of the configurations are the same. We just divide the previous answer by 7!. The answer is
10!
3!7!
Another way to solve this is to try picking 7 people to get gifts, out of 10 possible people. So the answer
is
10!
10
=
7
3!7!
5. If 8 identical blackboards are to be divided among 4 schools, how many divisions are possible? How
many if each school must receive at least 1 blackboard?
Solution:
n1 Number
n2 Number
n3 Number
n4 Number
of
of
of
of
blackboards
blackboards
blackboards
blackboards
given
given
given
given
to
to
to
to
school
school
school
school
1,
2,
3,
4.
n1 + n2 + n3 + n4 = 8.
There is a special formula for this. If each school must receive at least 1 blackboard
7!
7
8−1
=
=
= 35ways.
3
4−1
3!4!
If any school may receive no blackboard,
11!
8+4−1
11
=
=
= 165ways.
4−1
3
3!8!
6. In how many ways can r objects be selected from a set of n objects if the order of selection is considered
relevant?
Solution:
n
n−1
n−r+1
ways to choose the 1st object,
ways to choose the 2nd object,
...
ways to choose the rth object.
7. Determine the number of vectors (x1 , ..., xn ) such that each xi is either 0 or 1 and
n
X
i=0
2
xi ≥ k
Solution:
If we did not have the restriction on the sum, the answer would be 2n . However, with this restriction,
it means we need at least k 1s in the list, so we may see k ones, k + 1 ones, k + 2 ones, ...
Pn
n
→
ways this can happen,
k ones,
( i=1 xi = k)
k
Pn
n
k + 1 ones, ( i=1 xi = k + 1) →
ways this can happen,
k+1
... Pn
n
→
= 1 way this can happen.
n ones,
( i=1 xi = n)
n
The answer is
n X
n
i
i=k
8. A system is comprised of 5 components, each of which is either working or failed. Consider an experiment
that consists of observing the status of each component, and let the outcome of the experiment be given
by the vector (x1 , x2 , x3 , x4 , x5 ), where xi is equal to 1 if the component i is working and is equal to 0
if component i is failed.
(a) How many outcomes are in the sample space of this experiment?
(b) Suppose that the system will work if components 1 and 2 are both working, or if components 3
and 4 are both working, or if components 1 , 3 and 5 are all working. Let W be the event that the
system will work. Specify all the outcomes in W .
(c) Let A be the event that components 4 and 5 are both failed. How many outcomes are contained
in the event A?
(d) Write out all the outcomes in the event AW .
Solution:
(a) 25 = 32. I can list the sample space too:
(0, 0, 0, 0, 0)
(0, 0, 1, 0, 0)
(0, 1, 0, 0, 0)
(0, 1, 1, 0, 0)
(1, 0, 0, 0, 0)
(1, 0, 1, 0, 0)
(1, 1, 0, 0, 0)
(1, 1, 1, 0, 0)
(0, 0, 0, 0, 1)
(0, 0, 1, 0, 1)
(0, 1, 0, 0, 1)
(0, 1, 1, 0, 1)
(1, 0, 0, 0, 1)
(1, 0, 1, 0, 1)
(1, 1, 0, 0, 1)
(1, 1, 1, 0, 1)
(0, 0, 0, 1, 0)
(0, 0, 1, 1, 0)
(0, 1, 0, 1, 0)
(0, 1, 1, 1, 0)
(1, 0, 0, 1, 0)
(1, 0, 1, 1, 0)
(1, 1, 0, 1, 0)
(1, 1, 1, 1, 0)
(0, 0, 0, 1, 1)
(0, 0, 1, 1, 1)
(0, 1, 0, 1, 1)
(0, 1, 1, 1, 1)
(1, 0, 0, 1, 1)
(1, 0, 1, 1, 1)
(1, 1, 0, 1, 1)
(1, 1, 1, 1, 1)
(b) Let us define the events:
W1
W2
W3
=
=
=
event that components 1 and 2 are both working,
event that components 3 and 4 are both working,
event that components 1, 3 and 5 are all working.
P r [W ] = P r [W1 ∪ W2 ∪ W3 ]
= P r [W1 ] + P r [W2 ] + P r [W3 ] − P r [W1 W2 ] − P r [W1 W3 ] − P r [W2 W3 ] + 2P r [W1 W2 W3 ] .
3
To find P r [W1 ] note that the first two bits must be 1, but the remaining 3 can be 1 or 0. So there
are 23 ways to do this.
P r [W1 ] =
P r [W2 ] =
P r [W3 ] =
W1 W2
W1 W3
W2 W3
W1 W2 W3
=
=
=
=
event
event
event
event
that
that
that
that
components
components
components
components
1,
1,
1,
1,
2,
2,
3,
2,
23
1
=
25
4
3
2
1
=
25
4
2
1
2
=
25
8
3 and 4 are all working
3 and 5 are all working
4 and 5 are all working
3, 4 and 5 are all working
P r [W1 W2 ] =
P r [W1 W3 ] =
P r [W2 W3 ] =
P r [W1 W2 W3 ] =
21
25
21
25
21
25
1
25
1
16
1
=
16
1
=
16
1
=
32
=
P r [W ] = P r [W1 ∪ W2 ∪ W3 ]
= P r [W1 ] + P r [W2 ] + P r [W3 ] − P r [W1 W2 ] − P r [W1 W3 ] − P r [W2 W3 ] + 2P r [W1 W2 W3 ]
1 1 1
1
1
1
1
1
=
+ + −
−
−
+2×
=
4 4 8 16 16 16
32
2
Easy way: find the elements above where the machine would work:
(0, 0, 0, 0, 0)
(0, 0, 1, 0, 0)
(0, 1, 0, 0, 0)
(0, 1, 1, 0, 0)
(1, 0, 0, 0, 0)
(1, 0, 1, 0, 0)
(1 , 1 , 0 , 0 , 0 )
(1 , 1 , 1 , 0 , 0 )
(0, 0, 0, 0, 1)
(0, 0, 1, 0, 1)
(0, 1, 0, 0, 1)
(0, 1, 1, 0, 1)
(1, 0, 0, 0, 1)
(1 , 0 , 1 , 0 , 1 )
(1 , 1 , 0 , 0 , 1 )
(1 , 1 , 1 , 0 , 1 )
(0, 0, 0, 1, 0)
(0 , 0 , 1 , 1 , 0 )
(0, 1, 0, 1, 0)
(0 , 1 , 1 , 1 , 0 )
(1, 0, 0, 1, 0)
(1 , 0 , 1 , 1 , 0 )
(1 , 1 , 0 , 1 , 0 )
(1 , 1 , 1 , 1 , 0 )
(0, 0, 0, 1, 1)
(0 , 0 , 1 , 1 , 1 )
(0, 1, 0, 1, 1)
(0 , 1 , 1 , 1 , 1 )
(1, 0, 0, 1, 1)
(1 , 0 , 1 , 1 , 1 )
(1 , 1 , 0 , 1 , 1 )
(1 , 1 , 1 , 1 , 1 )
16
1
=
32
2
(c) If 4 and 5 are not working, the corresponding variables are 0. The other variables can be either 0
or 1. The number of ways this can happen (the number of outcomes in event A) is:
P r [W ] =
23 = 8.
(d) AW is the event that the system is working and elements 4 and 5 are not working. If 4 and 5 are
not working, events W2 and W3 above can’t be true. Thus only W1 defined above can be true,
AW = AW1 = event that components 1 and 2 are both working, 4 and 5 are not working.
In this case only component 3 may be 0 or 1. The outcomes in AW are (1, 1, 0, 0, 0) and (1, 1, 1, 0, 0).
4
9. Suppose that A and B are mutually exclusive events for which P (A) = .3 and P (B) = .5. What is the
probability that
(a) either A or B occurs?
(b) A occurs but B does not?
(c) both A and B occur?
Solution:
P (A) = 0.3
P (B) = 0.5
“A and B are mutually exclusive” means AB = ∅, and P r(AB) = 0.
(a) P r(A ∪ B) = P (A) + P (B) − P (AB) = P (A) + P (B) = 0.3 + 0.5 = 0.8.
(b) P r (B c A) = P r (B c |A) P r (A) According to the question, if A occurs, then it is certain that B does
not occur. Thus P r (B c |A) = 1. Finally, P r (B c A) = P r (B c |A) P r (A) = 1 × P (A) = 0.3.
(c) P r(AB) = 0.
10. An elementary school is offering 3 language classes: one in Spanish, on in French, and one in German.
The classes are open to any of the 100 students in the school. There are 28 students in the Spanish
class, 26 in the French class, and 16 in the German class. The 12 students that are in both Spanish
and French, 4 that are in both Spanish and German, and 6 that are in both French and German. In
addition, there are 2 students taking all 3 classes.
(a) If a student is chosen randomly, what is the probability that he or she is not in any of the language
classes?
(b) If a student is chosen randomly, what is the probability that he or she is taking exactly one of the
language classes?
(c) If 2 students are chosen randomly, what is the probability that at least 1 is taking a language class?
Solution:
(a) Number of students taking at least one language class = Spanish ∪ French ∪ German = Spanish +
French + German - (Spanish and German but not French) - (Spanish and French but not German)
- (French and German but not French) - (Spanish and German and French) = 28 + 26 + 16 (12-2) - (4-2) - (6-2) - 2 = 52 students take at least one language class.
Therefore 100 - 52 = 48 students take no language classes. So the probability that the student is
not taking any language class is 48/100 = 0.48.
(b)
Number of students taking exactly one language class
= Spanish + French + German
−2 × (Spanish and German but not French)
−2 × (Spanish and French but not German)
−2 × (French and German but not French)
−2 × (Spanish and German and French)
= 28 + 26 + 16 − 2 × (12 − 2) − 2 × (4 − 2) − 2 × (6 − 2) − 2 × 2
= 28 + 26 + 16 − 20 − 4 − 8 − 4 = 34
34 students take exactly one language class.
5
(c) Pr(at least one of the two is taking a language class) = 1-Pr(none of the two are taking a language
48 47
class) = 1 − 100
99 = 0.772.
11. Sixty percent of the students at a certain school wear neither a ring nor a necklace. Twenty percent
wear a ring and 30 percent a necklace. If one of the students is chosen randomly, what is the probability
that this student is wearing
(a) a ring or a necklace?
(b) a ring and a necklace?
Solution:
(a) Pr(ring or necklace) = 1 - Pr(no ring and no necklace) = 1 - 0.6 = 0.4.
(b) Pr(ring and necklace) = Pr(ring) + Pr(necklace) - Pr(ring and necklace) = 0.2 + 0.3 - 0.4 = 0.1.
12. A small community organization consists of 20 families, of which 4 have one child, 8 have two children,
5 have three children, 2 have four children, and 1 has five children.
(a) If one of these families is chosen at random, what is the probability it has i children, i = 1, 2, 3, 4, 5?
(b) If one of the children is randomly chosen, what is the probability that child comes from a family
having i children, i = 1, 2, 3, 4, 5?
Solution:
(a)
P r(i = 1) =
P r(i = 2) =
P r(i = 3) =
P r(i = 4) =
P r(i = 5) =
4
20
8
20
5
20
2
20
1
20
= 0.2
= 0.4
= 0.25
= 0.1
= 0.05
(b) Total number of children = 4 × 1 + 8 × 2 + 5 × 3 + 2 × 4 + 1 × 5 = 48.
P r(i = 1) =
P r(i = 2) =
P r(i = 3) =
P r(i = 4) =
P r(i = 5) =
4×1
48
8×2
48
5×3
48
2×4
48
1×5
48
=
=
=
=
=
4
48
16
48
15
48
8
48
5
48
13. An urn contains 3 red and 7 black balls. Players A and B withdraw balls from the urn consecutively
until a red ball is selected. Find the probability that A selects the red ball. (A draws the first ball, then
B, and so on. There is now replacement of the balls drawn.)
Solution:
6
who chooses
possible result
1)
2)
3)
4)
5)
6)
7)
8)
A
R
B
B
B
B
B
B
B
B
A
B
A
B
A
B
R
B
B
B
B
B
B
R
B
B
B
B
B
R
B
B
B
B
R
B
B
B
R
B
B
R
B
R
P r(A gets red ball)
= P r(red on 1st pull)
+ P r(black on 1st pull, black on 2nd pull, and red on 3rd pull)
+ P r(black on 1st pull, black on 2nd pull, black on 3rd pull, black on 4th pulland red on 5th pull)
+ P r (black on 1st pull, black on 2nd pull, black on 3rd pull, black on 4th pull, black on 5th pull,
=
=
black on 6th pulland red on 7th pull)
3
7
6 3
7
6 5 4 3
7
6 5 4 3 2 3
+
× × +
× × × × +
× × × × × ×
10 10 9 8 10 9 8 7 6 10 9 8 7 6 5 4
7
= 0.5833.
12
14. An urn contains 5 red, 6 blue, and 8 green balls. If a set of 3 balls is randomly selected, what is the
probability that each of the balls will be (a) of the same color? (b) of different colors? Repeat under the
assumption that whenever a ball is selected, its color is noted and it is then replaced in the urn before
the next selection. This is known as sampling with replacement.
Solution:
(a) Set of 3 balls of same color:
P r(All same color) = P r(All red) + P r(All blue) + P r(All green)
4
3
6
5
4
8
7
6
5
×
×
+
×
×
+
×
×
= 0.08873.
=
19 18 17 19 18 17 19 18 17
(b) Set of all different colors:
P r(all different colors) = P r(one red) + P r(one blue) + P r(one green)
5
6
8
=
×
×
× 3! = 0.24762
19 18 17
(a) Set of all 3 balls of same color sampling with replacement
P r(All same color) = P r(All red) + P r(All blue) + P r(All green)
5
5
5
6
6
6
8
8
8
×
×
+
×
×
+
×
×
= 0.12434.
=
19 19 19 19 19 19 19 19 19
(b) Set of all 3 balls of different colors sampling with replacement
P r(all different colors) = P r(one red) + P r(one blue) + P r(one green)
5
6
8
=
×
×
× 3! = 0.20994.
19 19 19
7
15. A forest contains 20 elk, of which 5 are captured, tagged, and then released. A certain time later, 4 of
the 20 elk are captured. What is the probability that 2 of these 4 have been tagged? What assumptions
are you making?
Solution:
5
4
15 14
P r(2 of 4 are tagged) =
×
×
×
×
20 19 18 17
4
2
=
7
= 0.184.
38
Assumption: any deer, whether tagged or untagged, is equally likely to be caught.
16. Five people designated as A, B, C, D, E, are arranged in linear order. Assuming that each possible
order is equally likely, what is the probability that
(a) there is exactly one person between A and B?
(b) there are exactly two people between A and B?
(c) there are exactly three people between A and B?
Solution:
(a) Finding the total number of arrangements where there is exactly 1 person between A and B.
1) A C B D E 3! ways A is on 1 and B is on 3.
2) B C A D E 3! ways B is on 1 and A is on 3.
3) D A C B E 3! ways A is on 2 and B is on 4.
4) D B C A E 3! ways B is on 2 and A is on 4.
5) D E A C B 3! ways A is on 3 and B is on 5.
6) D E B C A 3! ways B is on 3 and A is on 5.
Total number of arrangements is 3! × 6 = 36.
P r(There is exactly 1 person betwen A and B) =
36
36
=
= 0.3
5!
120
(b) Finding the total number of arrangements where there are exactly 2 people between A and B:
1) A C D B E 3! ways A is on 1 and B is on 4.
2) B C D A E 3! ways B is on 1 and A is on 4.
3) D A C E B 3! ways A is on 2 and B is on 5.
4) D B C E A 3! ways B is on 2 and A is on 5.
Total number of arrangements is 3! × 4 = 24.
P r(There are exactly 2 people betwen A and B) =
24
24
=
= 0.2
5!
120
(c) Finding the total number of arrangements where there are exactly 3 people between A and B:
1) A C D E B 3! ways A is on 1 and B is on 5.
2) B C D E A 3! ways B is on 1 and A is on 5.
Total number of arrangements is 3! × 2 = 12.
P r(There are exactly 3 people betwen A and B) =
12
12
=
= 0.1
5!
120
17. Let E, F , and G be three events. Find expressions for the events so that, of E, F , and G,
(a) only E occurs;
(b) both E and G but not F occur;
8
(c) at least one of the events occurs;
(d) at least two of the events occur;
(e) all three events occur;
(f) none of the events occur;
(g) at most one of the events occurs;
(h) at most two of the events occur;
(i) exactly two of the events occur;
(j) at most three of the events occur.
(a) only E occurs = EF c Gc ;
(b) both E and G but not F occur = EF Gc ;
(c) at least one of the events occurs = E ∪ F ∪ G;
(d) at least two of the events occur = EF ∪ F G ∪ EG;
(e) all three events occur = EF G;
(f) none of the events occur = E c F c Gc ;
(g) at most one of the events occurs = EF c Gc ∪ E c F Gc ∪ E c F c G ∪ E c F c Gc = (EF ∪ F G ∪ EG)c ;
(h) at most two of the events occur = (EF G)c ;
(i) exactly two of the events occur EF Gc ∪ EF c G ∪ E c F G;
(j) at most three of the events occur S (sample space).
18. Find the simplest expression for the following events:
(a) (E ∪ F ) (E ∪ F c );
(b) (E ∪ F ) (E c ∪ F ) (E ∪ F c );
(c) (E ∪ F ) (F ∪ G).
Solution:
(a) (E ∪ F ) (E ∪ F c ) = (E ∪ F ) E ∪ (E ∪ F ) F c = (EE ∪ F E) ∪ (EF c ∪ F F c ) = E ∪ EF c = E;
(b) (E ∪ F ) (E c ∪ F ) (E ∪ F c ); = ((E ∪ F ) E c ∪ (E ∪ F ) F ) (E ∪ F c ) = (EE c ∪ F E c ∪ EF ∪ F F ) (E ∪ F c )
= (F E c ∪ F E ∪ F F ) (E ∪ F c ) = F (E ∪ F c ) = F E ∪ F F c = F E
(c) (E ∪ F ) (F ∪ G) = E (F ∪ G) ∪ F (F ∪ G) = EF ∪ EG ∪ F F ∪ F G = F ∪ EG
19. Prove that P (E ∪ F ∪ G) = P (E) + P (F ) + P (G) − P (E c F G) − P (EF c G) − P (EF Gc ) − 2P (EF G).
Solution:
P (E ∪ F ∪ G) = P (E ∪ F ) + P (G) − P ((E ∪ F )G)
= P (E ∪ F ) + P (G) − P (EG ∪ F G)
= P (E) + P (F ) − P (EF ) + P (G) − (P (EG) + P (F G) − P (EGF G))
= P (E) + P (F ) + P (G) − P (EF ) − P (EG) − P (F G) + P (EF G)
Now we can expand some of these terms:
P (EF ) = P (EF G ∪ EF GC ) = P (EF G) + P (EF Gc )sinceP (EF GEF Gc ) = 0,
P (EG) = P (EF G ∪ EF c G) = P (EF G) + P (EF c G)sinceP (EF GEF c G) = 0,
P (F G) = P (EF G ∪ E c F G) = P (EF G) + P (E c F G)sinceP (EF GE c F G) = 0.
9
Finally,
P (E ∪ F ∪ G) = P (E) + P (F ) + P (G) − (P (EF G) + P (EF Gc ))
− (P (EF G) + P (EF c G)) − (P (EF G) + P (E c F G)) + P (EF G)
= P (E) + P (F ) + P (G) − P (EF Gc ) − P (EF c G) − P (E c F G) − 3P (EF G) + P (EF G)
= P (E) + P (F ) + P (G) − P (EF Gc ) − P (EF c G) − P (E c F G) − 2P (EF G)
20. If P (E) = .9 and P (F ) = .8, show that P (EF ) ≥ .7. In general, prove Bonferroni’s inequality, namely,
P (EF ) ≥ P (E) + P (F ) − 1
Solution:
We know that P (E ∪ F ) = P (E) + P (F ) − P (EF ). Since all probabilities must be less than 1,
P (E) + P (F ) − P (EF ) ≤ 1
P (E) + P (F ) − 1 ≤ P (EF )
P (EF ) ≥ P (E) + P (F ) − 1
P (EF ) ≥ 0.9 + 0.8 − 1 = 0.7.
21. What is the probability that at least one of a pair of fair dice lands on 6, given that the sum of the dice
is i, i = 2, 3, ..., 12?
Solution:
If i < 7, the answer is zero, since any number (on one dice) plus 6 (on the other dice) is greater than 6.
When we roll two dice, there are 6 × 6 possible answers.
• The following pairs of numbers yield a sum of 7: {(1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)} so the
probability of getting at least one 6 when the sum is 7 is 2/6.
• The following pairs of numbers yield a sum of 8: {(2, 6), (3, 5), (4, 4), (5, 3), (6, 2)} so the probability
of getting at least one 6 when the sum is 8 is 2/5.
• The following pairs of numbers yield a sum of 9: {(3, 6), (4, 5), (5, 4), (6, 3)} so the probability of
getting at least one 6 when the sum is 9 is 2/4.
• The following pairs of numbers yield a sum of 10: {(4, 6), (5, 5), (6, 4)} so the probability of getting
at least one 6 when the sum is 10 is 2/3.
• The following pairs of numbers yield a sum of 11: {(5, 6), (6, 5)} so the probability of getting at
least one 6 when the sum is 11 is 1.
• Only (6, 6) will yield a sum of 12, so {(5, 6), (6, 5)} so the probability of getting at least one 6 when
the sum is 12 is 1.
P r(one of the dice is 6|sum of the dice is N) =
10

0



1


3


2



 51
2
2


3



1




1


0
ifN ≤ 6
ifN = 7
ifN = 8
ifN = 9
ifN = 10
ifN = 11
ifN = 2
otherwise.
22. An urn contains 6 white and 9 black balls. If 4 balls are to be randomly selected without replacement,
what is the probability that the first 2 selected are white and the last 2 black?
Solution:
6
15
×
5
14
×
9
13
×
8
12
= 0.06592
23. Consider an urn containing 12 balls, of which 8 are white. A sample of size 4 is to be drawn with
replacement (without replacement). What is the conditional probability (in each case) that the first
and third balls drawn will be white given that the sample contains exactly 3 white balls?
Solution:
There are 8 white and 4 black balls.
W : Event that there are exactly 3 white balls
W = {(BW W W ), (W BW W ), (W W BW ), (W W W B)}
Z : Event that 3rd and 1st balls are white
Z = {(W BW B), (W BW W ), (W W W B), (W W W W )}
W Z = {(W BW W ), (W W W B)}
Without replacement:
Probabilities of these events happening are:
P r[W ] =
=
=
P r[Z] =
=
P r[W Z] =
=
4
8
7
6
8
4
7
6
8
7
4
6
8
7
6
4
×
×
× +
×
×
× +
×
×
× +
×
×
×
12 11 10 9 12 11 10 9 12 11 10 9 12 11 10 9
4
8
7
6
×
×
×
4×
12 11 10 9
0.45252
8
4
7
3
8
4
7
6
8
7
6
4
8
7
6
5
×
×
× +
×
×
× +
×
×
× +
×
×
×
12 11 10 9 12 11 10 9 12 11 10 9 12 11 10 9
0.42424
8
4
7
6
8
7
6
4
×
×
× +
×
×
×
12 11 10 9 12 11 10 9
0.22626
P r [Z|W ] =
P r [W Z]
0.22626
8×4×7×6+8×7×6×4
1
=
=
=
P r [W ]
0.45252
4×4×8×7×6
2
With replacement:
Probabilities of these events happening are:
3
8
4
×
= 0.39506
P r[W ] = 4 ×
12
12
2 2
3 1 4
8
4
8
4
8
P r[Z] =
+2×
+
= 0.44444
12
12
12
12
12
3 8
4
P r[W Z] = 2 ×
= 0.19750
12
12
P r [Z|W ] =
P r [W Z]
0.19750
2 × 4 × 83
1
=
=
=
3
P r [W ]
0.39506
4×8 ×4
2
11
24. An ectopic pregnancy is twice as likely to develop when the pregnant woman is a smoker as it is when
she is a nonsmoker. If 32 percent of women of childbearing age are smokers, what percentage of women
having ectopic pregnancies are smokers?
Solution:
E : Event that the woman has an ectopic pregnancy.
S : Event that the woman is a smoker.
We are given:
P r[S] = 0.32
P r [E|S] = 2 × P r [E|S c ]
We are asked to find P r[S|E].
P r[S|E] =
P r[ES]
P r[E|S]P r[S]
=
P r[E]
P r[E]
Also known:
P r[E] = P r[E|S]P r[S] + P r [E|S c ] P r [S c ]
Therefore:
P r[S|E] =
=
P r[E|S]P r[S]
2 × P r[E|S c ]P r[S]
=
P r[E|S]P r[S] + P r [E|S c ] P r [S c ]
2 × P r[E|S c ]P r[S] + P r [E|S c ] P r [S c ]
2P r[S]
2 × 0.32
=
= 0.48484
c
2P r[S] + P r [S ]
2 × 0.32 + (1 − 0.32)
25. A total of 500 married working couples were polled about their annual salaries, with the following
information resulting:
Wife
Less than $25,000
More than $25,000
Husband
Less than $25,000 More than $25,000
212
198
36
54
For instance, in 36 of the couples, the wife earned more and the husband earned less than $25,000. If
one of the couples is randomly chosen, what is
(a) the probability that the husband earns less than $25,000?
(b) the conditional probability that the wife earns more than $25,000 given that the husband earns
more than this amount?
(c) the conditional probability that the wife earns more than $25,000 given that the husband earns
less than this amount?
Solution:
(a) the probability that the husband earns less than $25,000?
Total number of couples = 212+36+198+54=500.
Number of couples where the husband earns less than $25,000 = 212+36=248.
P r[husband earns less than $25,000] =
12
248
= 0.496
500
(b) the conditional probability that the wife earns more than $25,000 given that the husband earns
more than this amount?
P r[W > $25, 000|H > $25, 000] =
P r[W > $25, 000, H > $25, 000]
54
=
= 0.21428
P r[H > $25, 000]
54 + 198
(c) the conditional probability that the wife earns more than $25,000 given that the husband earns
less than this amount?
P r[W > $25, 000|H < $25, 000] =
36
P r[W > $25, 000, H < $25, 000]
=
= 0.14516
P r[H < $25, 000]
36 + 212
26. In a certain community, 36 percent of the families own a dog and 22 percent of the families that own a
dog also own a cat. In addition, 30 percent of the families own a cat. What is
(a) the probability that a randomly selected family owns both a dog and a cat?
(b) the conditional probability that a randomly selected family owns a dog given that it owns a cat?
Solution:
D : Event that the family owns a dog.
S : Event that the family owns a cat.
Given: P r[D] = 0.36, P r[C|D] = 0.22, P r[C] = 0.30.
(a) the probability that a randomly selected family owns both a dog and a cat?
P r[CD] = P r[C|D]P r[D] = 0.22 ∗ 0.36 = 0.0792
(b) the conditional probability that a randomly selected family owns a dog given that it owns a cat?
P r[D|C] =
P r[CD]
0.0792
=
= 0.264
P r[C]
0.30
27. Suppose that 5 percent of men and .25 percent of women are color blind. A color blind person is chosen
at random. What is the probability of this person being male? Assume that there are an equal number
of males and females. What if the population consisted of twice as many males as females?
Solution:
C : Event that someone is color blind.
M : Event that someone is male.
W : Event that someone is female.
If there are an equal number of men and women, P r[M ] = P r[W ] = 0.5.
P r[M |C] =
P r[C|M ]P r[M ]
0.05 × 0.5
0.05
P r[M C]
=
=
=
= 0.95238
P r[C]
P r[C|M ]P r[M ] + P r[C|W ]P r[W ]
0.05 × 0.5 + 0.0025 × 0.5
0.0525
If there are twice as many males as females, P r[M ] = 2P r[W ], and since P r[W ] + P r[M ] = 1, we find
P r[W ] = 1/3, P r[M ] = 2/3.
P r[M |C] =
0.05 × 23
P r[M C]
P r[C|M ]P r[M ]
=
=
P r[C]
P r[C|M ]P r[M ] + P r[C|W ]P r[W ]
0.05 × 23 + 0.0025 ×
13
1
3
=
0.10
= 0.97560
0.1025
28. Suppose that an insurance company classifies people into one of three classes: good risks, average risks,
and bad risks. The company’s records indicate that the probabilities that good-, average- and bad-risk
persons will be involved in an accident over a 1-year span are, respectively, .05, .15, and .30. If 20
percent of the population is a good risk, 50 percent an average risk, and 30 percent a bad risk, what
proportion of people have accidents in a fixed year? If policy holder A had no accidents in 1997, what
is the probability that he or she is a good or average risk?
Solution:
G : Event
V : Event
B : Event
A : Event
that
that
that
that
someone
someone
someone
someone
is good risk,
is average risk,
is bad risk,
has an accident in one year.
We are given P r[A|G] = 0.05, P r[A|V ] = 0.15, P r[A|B] = 0.30. P r[G] = 0.20, P r[V ] = 0.50, P r[B] =
0.30.
The probability that a randomly chosen person has an accident is:
P r[A] = P r[A|G]P r[G] + P r[A|V ]P r[V ] + P r[A|B]P r[B] = 0.05 × 0.2 + 0.15 × 0.50 + 0.30 × 0.30 = 0.175.
Therefore the proportion of people who have accidents is 17.5%.
Second question asks the probability that a person is good or average risk given that they had no
accidents in a given year. In math: P r[G ∪ V |Ac ].
P r[BAc ]
P r[Ac |B]P r[B]
=
1
−
Ac
Ac
(1 − 0.30) × 0.30
= 1−
= 1 − 0.25454 = 0.74545
1 − 0.175
P r[G ∪ V |Ac ] = 1 − P r[B|Ac ] = 1 −
29. A ball is in any one of n boxes and is in the ith box with probability Pi . If the ball is in box i, a search
of that box will uncover it with probability αi . Show that the conditional probability that the ball is in
box j, given that a search of box i did not uncover it, is
Pj
1−αi Pi
(1−αi )Pi
1−αi Pi
if j 6= i
if j = i
Solution:
Bi : Event that ball is in box i,
Fi : Event that ball is found in box i,
P r [Fi |Bi ] = αi .
We are asked to find P r [Bj |Fic ].
P r [Bj |Fic ] =
P r [Fic |Bj ] P r [Bj ]
P r [Fic |Bj ] Pj
P r [Bj Fic ]
P
P
=
=
n
n
c
c
P r [Fic ]
k=1 P r [Fi |Bk ] P r [Bk ]
k=1 P r [Fi |Bk ] Pk
Probability of not finding a ball in a box if it isn’t in the box is 1. So,
1
if i 6= j
c
P r [Fi |Bj ] =
(1 − αi ) if i = j
and
n
X
k=1
P r [Fic |Bk ] Pk =
n
X
!
Pk
− Pi + (1 − αi ) Pi = 1 − Pi + Pi − αi Pi = 1 − αi Pi
k=1
14
if j 6= i:
P r [Fic |Bj ] Pj
Pj
P r [Bj |Fic ] = Pn
c |B ] P = 1 − α P .
P
r
[F
i i
k
k
k=1
i
if j = i:
P r [Fic |Bj ] Pj
(1 − αi ) Pi
P r [Bj |Fic ] = Pn
.
=
c
1 − αi Pi
k=1 P r [Fi |Bk ] Pk
30. Three dice are rolled. By assuming that each of the 63 = 216 possible outcomes is equally likely, find
the probabilities attached to the possible values that X can take on, where X is the sum of the 3 dice.
Solution:
If N < 3, P r[N ] = 0.
If N = 3, this can only happpen if all 3 dice land on 1, so
P r(N = 3) =
1 1 1
1
× × =
6 6 6
216
Look at all possible sums:
1+1+1
1+1+2
1+1+3
1+1+4
1+1+5
1+1+6
=
=
=
=
=
=
3
4
5
6
7
8
1+2+1
1+2+2
1+2+3
1+2+4
1+2+5
1+2+6
=
=
=
=
=
=
4
5
6
7
8
9
2+1+1
2+1+2
2+1+3
2+1+4
2+1+5
2+1+6
=
=
=
=
=
=
4
5
6
7
8
9
2+2+1
2+2+2
2+2+3
2+2+4
2+2+5
2+2+6
=
=
=
=
=
=
5
6
7
8
9
10
3+1+1
3+1+2
3+1+3
3+1+4
3+1+5
3+1+6
=
=
=
=
=
=
5
6
7
8
9
10
3+2+1
3+2+2
3+2+3
3+2+4
3+2+5
3+2+6
=
=
=
=
=
=
6
7
8
9
10
11
1+3+1
1+3+2
1+3+3
1+3+4
1+3+5
1+3+6
2+3+1
2+3+2
2+3+3
2+3+4
2+3+5
2+3+6
3+3+1
3+3+2
3+3+3
3+3+4
3+3+5
3+3+6
=
=
=
=
=
=
5
6
7
8
9
10
=
=
=
=
=
=
=
=
=
=
=
=
6
7
8
9
10
11
7
8
9
10
11
12
1+4+1
1+4+2
1+4+3
1+4+4
1+4+5
1+4+6
2+4+1
2+4+2
2+4+3
2+4+4
2+4+5
2+4+6
3+4+1
3+4+2
3+4+3
3+4+4
3+4+5
3+4+6
We can see a pattern emerge. Let’s put them in one big matrix:
15
=
=
=
=
=
=
6
7
8
9
10
11
=
=
=
=
=
=
=
=
=
=
=
=
7
8
9
10
11
12
8
9
10
11
12
13
1+5+1
1+5+2
1+5+3
1+5+4
1+5+5
1+5+6
2+5+1
2+5+2
2+5+3
2+5+4
2+5+5
2+5+6
3+5+1
3+5+2
3+5+3
3+5+4
3+5+5
3+5+6
=
=
=
=
=
=
1+6+1
1+6+2
1+6+3
1+6+4
1+6+5
1+6+6
7
8
9
10
11
12
=
=
=
=
=
=
=
=
=
=
=
=
8
9
10
11
12
13
9
10
11
12
13
14
2+6+1
2+6+2
2+6+3
2+6+4
2+6+5
2+6+6
3+6+1
3+6+2
3+6+3
3+6+4
3+6+5
3+6+6
=
=
=
=
=
=
8
9
10
11
12
13
=
=
=
=
=
=
=
=
=
=
=
=
9
10
11
12
13
14
10
11
12
13
14
15
3
4
5
6
7
8
4
5
6
7
8
9
5
6
7
8
9
10
6
7
8
9
10
11
7
8
9
10
11
12
8
9
10
11
12
13
4
5
6
7
8
9
5
6
7
8
9
10
6
7
8
9
10
11
7
8
9
10
11
12
8
9
10
11
12
13
9
10
11
12
13
14
5
6
7
8
9
10
6
7
8
9
10
11
7
8
9
10
11
12
8
9
10
11
12
13
9
10
11
12
13
14
10
11
12
13
14
15
6
7
8
9
10
11
7
8
9
10
11
12
8
9
10
11
12
13
9
10
11
12
13
14
10
11
12
13
14
15
11
12
13
14
15
16
7
8
9
10
11
12
8
9
10
11
12
13
9
10
11
12
13
14
10
11
12
13
14
15
11
12
13
14
15
16
12
13
14
15
16
17
8
9
10
11
12
13
9
10
11
12
13
14
10
11
12
13
14
15
11
12
13
14
15
16
12
13
14
15
16
17
13
14
15
16
17
18
Each block has a matrix structure known as a Toeplitz matrix. We can read out the probabilities using
16
this matrix:
P r[N ] =

























































1
216
3
216
3+2+1
6
216 = 216
4+3+2+1
10
= 216
216
15
5+4+3+2+1
= 216
216
6+5+4+3+2+1
21
= 216
216
5+6+5+4+3+2
25
= 216
216
4+5+6+5+4+3
27
= 216
216
3+4+5+6+5+4
27
= 216
216
2+3+4+5+6+5
25
= 216
216
21
1+2+3+4+5+6
= 216
216
1+2+3+4+5
15
= 216
216
1+2+3+4
10
= 216
216
1+2+3
6
216 = 216
1+2
3
216 = 216
1
1
216 = 216
if
if
if
if
if
if
if
if
if
if
if
if
if
if
if
if
N
N
N
N
N
N
N
N
N
N
N
N
N
N
N
N
=3
=4
=5
=6
=7
=8
=9
= 10
= 11
= 12
= 13
= 14
= 15
= 16
= 17
= 18
31. Five men and 5 women are ranked according to their scores on an examination. Assume that no two
scores are alike and all 10! possible rankings are equally likely. Let X denote the highest ranking
achieved by a woman. (For instance, X = 1 if the top-ranked person is female.) Find P {X = i},
i = 1, 2, 3, ..., 8, 9, 10.
Solution:
X = highest rank achieved by a woman.
Without any restrictions, there are 10! ways to rank everyone.
First, X = 1 with probability 0.5.
If X = 2, then the highest ranking person must be a man, and the second highest ranking person must
be a woman. We can choose 1 man from 5 to be highest ranking, and 1 woman from the five women to
be highest ranking. The remaining 8 people will rank among themselves in 8! ways, so:
5
5
8!
1
1
5 × 5 × 8!
25
P r(X = 2) =
=
=
= 0.27777
10!
10!
90
If X = 3, then the two highest ranking people must be male, and the third highest ranking person must
be a woman. We can choose 2 men from 5 to be highest ranking, and then we can arrange them in 2!
ways. We choose 1 woman from the five women to be highest ranking. The remaining 7 people will
rank among themselves in 7! ways, so:
5
5
× 2!
× 7!
5!
2
1
× 2 × 5 × 7!
100
P r(X = 3) =
= 3!2!
=
= 0.13888
10!
10!
720
If X = 4, then the three highest ranking people must be male, and the fourth highest ranking person
must be a woman. We can choose 3 men from 5 to be highest ranking, and then we can arrange them
in 3! ways. We choose 1 woman from the five women to be highest ranking. The remaining 6 people
will rank among themselves in 6! ways, so:
5
5
× 3!
× 6!
5!
3
1
× 3! × 5 × 6!
300
P r(X = 4) =
= 3!2!
=
= 0.05952
10!
10!
5040
17
Similarly,
P r(X = 5) =
5
4
× 4!
5
1
× 5!
=
10!
P r(X = 6) =
5!
4!1!
× 4! × 5 × 5!
600
=
= .01984
10!
30240
5!5!
120
=
= 0.00396
10!
30240
32. Four independent flips of a coin are made. Let X denote the number of heads obtained. Plot the
probability mass function of the random variable X − 2.
P r(X = 0) = 0.54 = 0.0625
4
P r(X = 1) =
0.54 = 0.25
1
4
P r(X = 2) =
0.54 = 0.375
2
4
0.54 = 0.25
P r(X = 3) =
3
4
0.54 = 0.0625
P r(X = 4) =
4
The pmf of X − 2 is:
18
0.4
0.35
0.3
0.25
0.2
0.15
0.1
0.05
0
−2
−1
19
0
1
2
33. If the distribution function of X is given by
F (b) =

0



1



 23
5
4
5
9
10







1
b<0
0≤b<1
1≤b<2
2≤b<3
3 ≤ b < 3.5
b ≥ 3.5.
calculate the probability mass function of X.
Solution:
f (x) =















1
2
1
10
1
5
1
10
1
10
0
b=0
b=1
b=2
b=3
b = 3.5
otherwise.
34. Four busses carrying 148 students from the same school arrive at a football stadium. The buses carry,
respectively, 40, 33, 25, and 50 students. One of the students is randomly selected. Let X denote the
number of students that were on the bus carrying the randomly selected student. One of the 4 bus
drivers is also randomly selected. Let Y denote the number of students on her bus.
(a) Which of E [X] or E [Y ] do you think is larger? Why?
(b) Compute E [X] and E [Y ].
Solution:
(a) Which of E [X] or E [Y ] do you think is larger? Why?
E[X] should be larger because of sampling bias.
(b) Compute E [X] and E [Y ].
E[X] = 40 ×
40
33
25
50
+ 33 ×
+ 25 ×
+ 50 ×
= 39.28378
148
148
148
148
E[Y ] = 40 ×
1
1
1
1
+ 33 × + 25 × + 50 × = 37.0
4
4
4
4
35. A box contains 5 red and 5 blue marbles. Two marbles are withdrawn randomly. If they are the same
color, then you win $1.10; if they are different colors then you win -$1.00 (That is, you lose $1.00.)
Calculate
(a) the expected value of the amount you win;
(b) the variance of the amount you win.
Solution:
(a) the expected value of the amount you win;
20
E[X] = 2P r(red, red) − 0.1 (P r(red, blue) + P r(blue, red)) − 2.2P r(blue, blue)
5×4
5×5
5×4
= 2
− 0.1 × 2 ×
− 2.2 ×
10 × 9
10 × 9
10 × 9
2
5
2
= 2 × − 0.1 × − 2.2 ×
9
9
9
4 − 0.5 − 4.4
=
= −0.1
9
(b) the variance of the amount you win.
V ar(X) = (2 − (−0.1))2 P r(red, red) + (−0.1 − (−0.1))2 (P r(red, blue) + P r(blue, red))
+(−2.2 − (−0.1))2 P r(blue, blue)
5×4
5×4
+ (−2.1)2 ×
= 1.92
10 × 9
10 × 9
= 1.78222
36. Suppose that, in flight, airplane engines will fail with probability 1 − p, independently from engine to
engine. If an airplane needs a majority of its engines operative to complete a successful flight, for what
values of p is a 5-engine plan preferable to a 3-engine plane?
Solution:
5 engine plane:
P r(all engines fail) = (1 − p)5
3 engine plane:
P r(all engines fail) = (1 − p)3
The 5 engine plane is less likely to crash if
(1 − p)5 < (1 − p)3
(1 − p)2 < 1
Since 1 − p is always less than 1, its square will be less than 1 also. So the 5 engine plane is always safer.
37. A communications channel transmits the digits 0 and 1. However, due to static, the digit transmitted
is incorrectly received with probability .2. Suppose that we want to transmit an important message
consisting of one binary digit. To reduce the chance of error, we will transmit 00000 instead of 0 and
11111 instead of 1. If the receiver of the message uses “majority” decoding, what is the probability that
the message will be wrong when decoded? What independence assumptions are you making?
Solution:
P r[wrong] = P r[between 3 and 5 incorrect bits)
5
5
5
3
2
4
1
=
0.2 × 0.8 +
0.2 × 0.8 +
0.25
3
4
5
5!
5!
=
0.23 × 0.82 +
0.24 × 0.81 + 0.25
3!2!
4!1!
= 10 × 0.23 × 0.82 + 5 × 0.24 × 0.81 + 0.25 = 0.05792
We are assuming that the probability that the bit is wrong does not depend on the probability that the
previous bit is wrong.
21
38. It is known that diskettes produced by a certain company will be defective with probability .01, independently of each other. The company sells the diskettes in packages of size 10 and offers a money-back
guarantee that at most 1 of the 10 diskettes in the package will be defective. The guarantee is that the
customer can return the entire package of diskettes if he or she finds more than one defective diskette
in it. If someone buys 3 packages, what is the probability that he or she will return exactly 1 of them?
Solution:
Pr(defective diskette) = 0.01.
P r(exactly 1 return) =
3
1
P r(more than 1 defective diskette)P r(1 or fewer defective diskettes)2
10
0.999 0.011 = 0.99573
1
P r(2 or more defective diskettes) = 1 − 0.99573 = 0.0427
P r(1 or fewer defective diskettes) = 0.9910 +
P r(exactly 1 return) = 3 × 0.0427 × 0.995732 = 0.01270
39. When coin 1 is flipped, it lands on heads with probability .4; when coin 2 is flipped, it lands on heads
with probability .7. One of these coins is randomly chosen and flipped 10 times.
(a) What is the probability that the coin lands on heads on exactly 7 of the 10 flips?
(b) Given that the first of these ten flips lands heads, what is the conditional probability that exactly
7 of the 10 flips lands on heads?
Solution:
7H : Event that coin lands on heads exactly 7 times,
C1 : Event that coin 1 is picked,
C2 : Event that coin 2 is picked.
(a) What is the probability that the coin lands on heads on exactly 7 of the 10 flips?
P r[7H] = P r[7H|C1]P r[C1] + P r[7H|C2]P r[C2]
10
10
7
3
=
0.4 0.6 × 0.5 +
0.77 0.33 × 0.5
7
7
10!
=
0.5 0.47 0.63 + 0.77 0.33 = 0.15420
7!3!
(b) Given that the first of these ten flips lands heads, what is the conditional probability that exactly
7 of the 10 flips lands on heads?
The fact that the first flip lands heads changes the probability of which coin was chosen.
H1 : Event that the coin lands heads the first time.
22
P r[C1|H1] =
=
P r[C2|H1] =
=
P r(H1|C1)P r(C1)
P r(H1|C1)P r(C1)
=
P r(H1)
P r(H1|C1)P r(C1) + P r(H1|C2)P r(C2)
0.4 × 0.5
4
=
0.4 × 0.5 + 0.7 × 0.5
11
P r(H1|C2)P r(C2)
P r(H1|C2)P r(C2)
=
P r(H1)
P r(H1|C1)P r(C1) + P r(H1|C2)P r(C2)
0.7 × 0.5
7
=
0.4 × 0.5 + 0.7 × 0.5
11
Therefore,
P r[7H] = P r[7H|C1, H1]P r[C1|H1] + P r[7H|C2, H1]P r[C2|H1]
7
4
9
9
+
0.76 0.33 ×
=
0.46 0.63 ×
6
6
11
11
10!
=
0.5 0.47 0.63 + 0.77 0.33 = 0.19676
7!3!
40. If you buy a lottery ticket in 50 lotteries, in each of which your chance of winning a prize is
is the (approximate) probability that you will win a prize
1
100 ,
what
(a) at least once?
(b) exactly once?
(c) at least twice?
Solution:
(a) at least once?
1 50
P r[at least one win] = 1 − P r[no wins] = 1 − 1 −
= 0.395
100
(b) exactly once?
P r[exactly one win] =
50
1
1
100
1 49
1−
= 0.3055
100
(c) at least twice?
P r[at least two wins] = 1 − P r[exactly one win] − P r[no wins] = 1 − 0.3055 − 0.9950 = 0.08950
41. Let X be a random variable having expected value µ and variance σ 2 . Find the expected value and
variance of
X −µ
Y =
σ
Solution:
X −µ
E[X] − µ
E[Y ] = E
=
= 0 = µY .
σ
σ
23
h
i
h
i
V ar(Y ) = E (Y − µY )2 = E (Y − 0)2 = E Y 2
h
i
"
"
#
2
2 #
2
E
(X
−
µ)
X −µ
(X − µ)
= E
=E
=
σ
σ2
σ2
=
V ar(X)
V ar(X)
σ2
=
= 2 =1
2
2
σ
σ
σ
42. Find V ar(X) if
P (X = a) = p = 1 − P (X = b)
Solution:
E[X] = pa + (1 − p)b = µ
h
i
V ar(X) = E (X − µ)2
= (a − pa − (1 − p)b)2 p + (b − pa − (1 − p)b)2 (1 − p)
= (a − b)2 (1 − p)2 p + p2 (b − a)2 (1 − p)
= (a − b)2 p(1 − p)((1 − p) + p)
= (a − b)2 p(1 − p)
24