Probability and Statistics Assignment: Exercises and Solutions

This maths assignment presents solutions to various probability and statistics problems. The assignment covers topics including probability distributions, expected value calculations, combinations, permutations, and binomial distributions. Several exercises involve calculating probabilities for different scenarios, such as the number of boys born, the sum of dice rolls, and card draws. The solutions demonstrate the application of formulas and concepts to solve these problems, including the use of probability trees and binomial distribution calculations. The document provides detailed step-by-step solutions for each problem, making it a valuable resource for students studying statistics and probability. The assignment also explores the use of combinations and permutations in different scenarios, such as card games and lottery tickets. The final section covers binomial distributions, including calculating probabilities for specific outcomes. The assignment aims to provide a comprehensive understanding of probability and statistics concepts.

Maths Assignment
Student Name:
Instructor Name:
Course Number:
25 April 2019

9.1 Exercises
1) Four children are born, and the number of boys noted.
Answer
x 0 1 2 3 4
P(x) .0625 .25 .375 .25 .0625
2) Two dice are rolled, and the total number of dots recorded
Answer
x 2 3 4 5 6 7 8 9 10 11 12
P(x) 1/36 2/36 3/36 4/36 5/36 6/36 5/36 4/36 3/36 2/36 1/36
3) Three cards are drawn from a deck. The number of Queens are counted
Answer
x 0 1 2 3
P(x) .782624 .204163 .013032 .000181
4) Two names are drawn from a hat, signifying who should go pick up pizza. Three of the names
are on the swim team and two are not. The number of swimmers selected is counted.
Answer
x 0 1 2
P(x) .1 .6 .3
5) Exercise 1; P ( x ≤ 2 )
Answer
P ( x ≤ 2 ) =P ( x=0 ) +P ( x=1 ) +P ( x=2 ) =0.0625+0.25+ 0.375=0.6875
6) Exercise 2; P ( x ≥ 11 )
Answer
P ( x ≥ 11 )=P ( x=11 ) + P ( x=12 )= 2
36 + 1
36 = 3
36 = 1
12 =0.0833
7) Exercise 3; P(at least one queen)

Answer
P ( at least 1 queen )=P ( x=1 ) +P ( x=2 ) +P ( x=3 )=0.204163+0.013032+0 .000181
¿ 0.217376
8) Exercise 4; P(fewer than two swimmers)
Answer
P ( fewer thantwo swimmers ) =P ( x <2 ) =P ( x=0 ) + P ( x=1 ) =0.1+0.6
¿ 0.7
9)
X 1 3 5 7
P(x) .1 .5 .2 .2
Answer
E ( x )=∑ xi P( xi )= ( 1∗0.1 ) + ( 3∗0.5 ) + ( 5∗0.2 )+ ( 7∗0.2 )=0.1+1.5+1.0+1.4=4.0
10)
y 0 15 30 40
P(y) .15 .20 .40 .25
Answer
E ( x )=∑ xi P( xi )= ( 0∗0.15 ) + ( 15∗0.20 ) + ( 30∗0.40 ) + ( 40∗0.25 )=0+3.0+ 12.0+10.0=25.0
11)
z 0 2 4 8 16
P(z) .21 .24 .21 .17 .17
Answer
E ( x ) =∑ xi P( xi )= ( 0∗0.21 ) + ( 2∗0.24 ) + ( 4∗0.21 ) + ( 8∗0.17 ) + ( 16∗0.17 ) =0+0.48+0.84+1.36+ 2.72=5.4
12)
x 5 10 15 20 25
P(x) .40 .30 .15 .10 .05

Answer
E ( x )=∑ xi P( xi )= ( 5∗0.40 ) + ( 10∗0.30 ) + ( 15∗0.15 ) + ( 20∗0.10 ) + ( 25∗0.05 )=2+ 3+2.25+2+1.25=10.5
13)
Answer
E ( x ) =∑ xi P( xi )= ( 1∗0.20 ) + ( 2∗0.30 ) + ( 3∗0.10 ) + ( 4∗0.40 ) =0.2+0.6+0.3+1.6=2.7
14)
Answer
E ( x )=∑ xi P( xi )= ( 2∗0.20 ) + ( 4∗0.30 ) + ( 6∗0.20 ) + ( 8∗0.10 ) + ( 10∗0.20 ) =0.4+1.2+1.2+0.8+ 2=5.6
15)
Answer
E ( x )=∑ xi P( xi )= ( 1∗0.30 ) + ( 2∗0.25 ) + ( 3∗0.20 ) + ( 4∗0.15 )+ ( 5∗0.10 )=0.3+0.5+ 0.6+0.6+0.5=2.5
16)
Answer
E ( x )=∑ xi P( xi )= ( 10∗0.20 ) + ( 20∗0.20 ) + ( 30∗0.20 ) + ( 40∗0.20 )+ ( 50∗0.20 )=2+4+6+8+10=30
17)
Answer
E ( x ) =∑ xi P( xi )= ( 1 ) ( 18
38 ) + ( −1 ) ( 20
38 )= 18
38 − 20
38 =−2
38 =−0.0526∨−5.26 cents
18)
Answer
E ( x )=∑ xi P( xi )= ( 1 ) ( 19
38 )+ (−1 ) ( 20
38 )= 19
38 − 20
38 =−1
38 =−0.0263∨−2.63 cents
19)
Answer

E ( x )=∑ xi P ( xi )= (−1 ) ( 999
1000 )+ ( 499 ) ( 1
1000 )=−0.999+0.499
¿−0.50∨−50 cents
20)
Answer
E ( x )=∑ xi P ( xi )= (3.20 ) ( 20
80 )+ ( 0 ) ( 60
80 )−$ 1=$ 0.8−$ 1
¿−0.20∨−20 cents
21)
Answer
E ( x )=∑ xi P ( xi )= ( 49999 ) ( 1
2000000 )+ ( 9999 ) ( 2
2000000 )+ (−1 ) (1999997
2000000 )=0.0249995+ 0.009999−0.9999985
¿−0.965∨−96.5 cents
22)
Answer
E ( x )=∑ xi P ( xi )= (24999 ) ( 1
20000 )+ ( 9999 ) ( 2
20000 )+ ( 4999 ) ( 3
20000 )+ (−5 ) ( 19996
20000 )=1.24995+0.9999+0.74985
¿−1.9993∨−199.93 cents
23)
Answer
E ( x )=∑ xi P ( xi )= (100000 ) ( 1
8504860 )+ ( 50000 ) ( 1
302500 )+ ( 10000 ) ( 1
282735 )+ ( 1000 ) ( 1
153560 )+ (100 ) ( 1
104560 )
¿ 0.0117 6+0.165289+ 0.03537 +0.00651+0.0009 56+0.00262−1.89
¿−1.667495∨−166.75 cents
24)
Answer
E ( x )=∑ xi P ( xi )= (100000 ) ( 1
8504860 )+ ( 50000 ) ( 1
302500 )+ ( 10000 ) ( 1
282735 )+ ( 1000 ) ( 1
153560 )+ (100 ) ( 1
104560 )
¿ 0.0117 6+0.165289+0.03537 +0.00651+0.0009 56+0.00262−47.25

¿−47.027495∨−4702.7495 cents
25)
Answer
E ( x ) =∑ xi P ( xi )= ( 0∗.2463 ) + ( 1∗.4173 ) + ( 2∗.2470 ) + ( 3∗.0650 ) + ( 4∗.0064 ) =0+0.4173+0.494 +0.195+0.0256
¿ 1.1319
26)
Answer
E ( x )=∑ xi P ( xi )= ( 0∗.0206 )+ (1∗.1209 ) + ( 2∗.2838 )+ ( 3∗.3332 ) + ( 4∗.1956 ) + ( 5∗.0459 )
¿ 0+0.1209+ 0.5676+0.9996+0.7824 +0.2295
¿ 2.7
27)
Answer
∑ pi =1then valid
∑ pi =0.01+0.09+0.25+0.45+ 0.05+0.20−0.05=1
But we can’t have negative probability hence the probability distribution is invalid
28)
Answer
∑ pi =1then valid
∑ pi =0.05+0.10+0.75+ 0.02+ 0.03+0.04+ 0.01=1
This probability distribution is valid
29)
Answer
∑ pi =1then valid
∑ pi =0.01+0.02+0.03+0.04 +0.05+0.85=1
This probability distribution is valid

9.2 Exercises
1) 4 P2
Answer
4 P2= 4∗3∗2∗1
2 =4∗3=12
2) 3!
Answer
3 !=3∗2∗1=6
3) 8C5
Answer
8 C 5= 8 !
( 8−5 ) !5 ! = 8 !
3 ! 5! = 8∗7∗6∗5 !
3∗2∗5 ! =8∗7=56
4) 7!
Answer
7 !=7∗6∗5∗4∗3∗2∗1=5040
5) 8P1
Answer
8 P1= 8∗7∗6∗5∗4∗3∗2∗1
1 =40320
6) 7C2
Answer
7 C 2= 7 !
( 7−2 ) ! 2! = 7 !
5 ! 2 != 7∗6∗5 !
5 !∗2∗1 =7∗6
2 =21
7) 4!
Answer
4 !=4∗3∗2∗1=24

8) 4P4
Answer
4 P4= 4∗3∗2∗1
4∗3∗2∗1 =1
9) 9C6
Answer
9 C 6= 9 !
( 9−6 ) ! 6! = 9 !
3!6 ! = 9∗8∗7∗6 !
3∗2∗6 ! =3∗4∗7=84
10) 8C2
Answer
8 C 2= 8 !
( 8−2 ) ! 2! = 8 !
6 ! 2 ! = 8∗7∗6 !
6 !∗2∗1 =4∗7=28
11) 13P3
Answer
13 P3= 13!
3! =1037836800=1.0378368 ×1010
12) 9P5
Answer
9 P5= 9 !
5 ! = 9∗8∗7∗6∗5 !
5! =9∗8∗7∗6=3024
13) 25P5
Answer
25 P5= 25 !
5 ! =1.2926 ×1023
14) 40P5
Answer

40 P5 = 40!
5 ! =6.79929 ×1045
15) 14P5
Answer
14 P5= 14 !
5 ! =7.26486× 108
16) 17P8
Answer
17 P8= 17 !
8 ! =8.82161 ×109
17) 18C5
Answer
18 C5 = 18 !
( 18−5 ) ! 5! = 18 !
13 ! 5 ! =8568
18) 32C9
Answer
32 C9 = 32!
( 32−9 ) ! 9 != 32 !
23! 9! =2.80488 ×107
19) 28C14
Answer
28 C14 = 28 !
( 28−14 ) ! 14 ! = 28 !
14 !14 !=4.01166× 107
20) 35C30
Answer
35 C30= 35 !
( 35−30 ) ! 30! = 35 !
5 !30 !=3.24632 ×105
9.3 Exercises
1) Exactly 1 of the sampled tiles is defective;
Answer

P ( X=x ) =e− λ λx
x !
P ( X=1 )= e−5 51
1 ! =0.03369
2) The batch is rejected (that is, at least 1 sampled tile is defective).
Answer
P ( X=x ) =e− λ λx
x !
P ( At least 1 defective )=P ( X ≥ 1 )=P ( X =1 )+ P ( X =2 ) + P ( X=3 )= e−5 51
1 ! + e−5 52
2 ! + e−5 53
3 ! =0.03369+0.0842
3) 1
Answer
P ( X=x ) =e− λ λx
x !
P ( X=1 ) = e−3 31
1 ! =0.149361
4) 2
Answer
P ( X=x ) =e− λ λx
x !
P ( X=2 )= e−3 32
2 ! =0.224042
5) 3
Answer
P ( X=x ) =e− λ λx
x !
P ( X=3 )= e−3 33
3 ! =0.224042
6) 5
Answer
P ( X=x ) =e− λ λx
x !
P ( X=5 ) = e−3 35
5 ! =0.100819
7) All $100 tickets
Answer
Total number of tickets = 10+12+20+200=242

P ( all $ 100 ) =
10
242∗9
241 ∗8
240 =0.00005144
8) All $50 tickets
Answer
Total number of tickets = 10+12+20+200=242
P ( all $ 5 0 ) =
12
242∗11
241 ∗10
240 =0.000094304
9) Exactly two $25 tickets and no other money winners
Answer
P ( Exactly two $ 25 tickets∧no other money winners ) =
20
242∗19
241 ∗200
240 =0.00543
10) One ticket of each money amount
Answer
P ( Oneticket of each money amount )=
20
242∗12
241 ∗10
240 =0.00017146 2
11) No tickets with money
Answer
P ( Notickets with money ) =
200
242∗199
241 ∗198
240 =0.562995
12) At least one money ticket
Answer
Let $100 be A, $50 be B, $25 be C and no money be D
P ( At least one money ticket )=P ( ADD )+ P ( BDD ) + P ( CDD ) + P ( ABD ) + P ( ACD ) + P ( BCD )=( 10
242∗200
241 ∗1
240
13) How many 2-card hands are possible?
Answer
Possible ways=13∗13=169
14) 2 kings

Answer
P ( 2 kings ) = 4 C2
52C2
=0.004525
15) No deuces (2’s)
Answer
P ( Nodeuces ) =1− 4 C2
52 C2
=1−0.004525=0.995475
16) 2 face cards
Answer
P ( 2 face cards ) = 2∗4 C2
52C2
=2∗0.004525=0.00905
17) Different suits
Answer
P ( different suits ) = 4 C2
52 C5
= 23
128 =0.1796875
18) At least 1 black card
Answer
P ( at least 1 black card )= 2
52 = 1
26 =0.0385
19) No more than 1 diamond
Answer
P ( Nomore than 1 diamond ) = 19
34 + 13
52 =0.5588+0.25=0.8088
20) Discuss the relative merits of using probability trees versus combinations to solve
probability problems. When would each approach be most appropriate?
Answer
The relative merit of using probability trees versus combinations is that the fact that
probability trees are capable of assigning specific values to a problem, decisions, and
outcomes of each decision. Through this assignment of specific values, there is a
reduction in ambiguity in decision-making.
Probability trees are most appropriate when we don’t have many cases to be considered
while combinations is best when we are considering several cases.
9.4 Exercises
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