7.1 Probability Distribution pg 374 # 1 – 5, 6ab, 8 – 11
P(x) = 1/n
E(X) = ∑𝑛𝑖 𝑥𝑖 𝑃(𝑥𝑖 )
1.
a) discrete
b) continuous
c) continuous
2.
a) uniform – each ticket is worth 1
b) not uniform. Selector has a personal preference.
c) uniform – equal chance of heads or tales
d) uniform – random so equal
e) not uniform – guess is bias due to looks
f) uniform – well shuffled makes it random
g) no uniform – 1 two, 3 fours
3.
a) 5(0.3) + 10(0.25) + 15(0.45)
= 1.5 + 2.5 + 6.75
= 10.75
b) 1000(.25) + 100000(.25) +1000000(.25) + 10000000(.25)
= 250 + 25000 + 250000 + 2500000
= 2,775,250
c) 1(1/6) + 2(1/5) + 3(1/4) + 4(1/3) + 5(1/20)
= 1/6 + 2/5 + 3/4 + 4/3 + 5/20
= 10/60 + 24/60 + 45/60 + 80/60 + 15/60
= 174/60 = 87/30 = 29/10
d) discrete
e) continuous
f) discrete
4.
a) 1/2
b) 1(1/8) + 2(1/8) + 3(1/8) + 4(1/8) + 5(1/8) + 6(1/8) + 7(1/8) + 8(1/8)
= (1+2+3+4+5+6+7+8)(1/8)
= 36(1/8)
= 36/8 = 18/4 = 9/2
5.
a) Yes. Randomly selected ignoring private numbers
b) In your exchange, only the last four numbers are different. Therefore 10^4 = 10000
1/10000
c) 10/10000 = 1/1000
6.
a)
X
P(X)
2
1/36
3
2/36
4
3/36
5
4/36
6
5/36
7
6/36
8
5/36
9
4/36
10
3/36
11
2/36
b)
= 2(1/36)+3(2/36) +4( 3/36) +5( 4/36) +6( 5/36) +7( 6/36) +8( 5/36) +9( 4/36) +10( 3/36) +11( 2/36) +12( 1/36)
= 2/36 + 6/36 + 12/36 + 20/36 + 30/36 + 42/36 + 40/36 + 36/36 + 30/36 + 22/36 + 12/36
= 252/36 = 126/18 = 63/9 = 7
8.
a) 1+1+5 = 7
P(win) = 7/2000000
b) 1000000(1/2000000) + 25000(1/2000000) + 1000(5/2000000)
= 1000000/2000000 + 25000/2000000 + 5000/2000000
= 1030000/2000000 = $0.515
$0.515 – 2.00
= $-1.485
12
1/36
9.
X
Points
Probability
1
-3
1/6
2
4
1/6
3
-9
1/6
4
8
1/6
5
-15
1/6
6
12
1/6
a) = -3(1/6) + 4(1/6) + -9(1/6) + 8(1/6) + -15(1/6) + 12(1/6)
= (-1 + 4 – 9 + 8 – 15 + 12)(1/6)
= -3(1/6)
= -3/6 = -1/2 = -0.5
b) no, expected value does not equal zero.
10.
P(E) = 1000000(1/2000000) + 50000(5/2000000) + 1000(10/2000000) + 50(50/2000000)
= 1000000/2000000 + 250000/2000000 + 10000/2000000 + 2500/2000000
= 1262500/2000000
= $0.63
40% profit, multiply by 1.4
0.63 * 1.4
= 0.88
11. BB, BG, GB, GG
0(1/4) + 1(1/4) + 1(1/4) + 2(1/4)
= 1/4 + 1/4 + 2/4
= 4/4 = 1
7.2 Binomial Distributions pg 385 # 1 – 3, 5, 7, 8ab
Binomial Distribution
P(x) = nCxpxqn-x
E(X) = np
1.
a) yes, 3 or not a three
b) no, player 1 hit or miss, player 2 hit or miss. More than two options
c) no, how can you tell who had? Did it keep infection off?
d) yes, good or bad
2.
a) p = 0.2, then q = 0.8
X
0
1
2
3
4
5
P(X)
0
5-0
= 1(1)(.32768) = .32768
5C0(0.2) (0.8)
1
5-1
= 5(0.2)(.4096) = .4096
5C1(0.2) (0.8)
2
5-2
= 10(.04)(.512) = .2048
5C2(0.2) (0.8)
3
5-3
= 10(.008)(.64) = .00512
5C3(0.2) (0.8)
4
5-4
C
(0.2)
(0.8)
= 5(.0016)(.8) = .0064
5 4
5
5-5
= 1(.00032)(1) = .00032
5C5(0.2) (0.8)
b) p = .5, then q = 0.5
X
0
1
2
3
4
5
6
7
8
P(X)
0
8-0
= 1(1)(. 00390625) = . 00390625
8C0(0.5) (0.5)
1
8-1
= 8(0.5)(. 0078125) = .03125
8C1(0.5) (0.5)
2
8-2
= 28(.25)(. 015625) = .02048
8C2(0.5) (0.5)
3
8-3
= 56(.125)(. 03125) = .000512
8C3(0.5) (0.5)
4
8-4
= 70(.0625)(. 0625) = .00128
8C4(0.5) (0.5)
5
8-5
C
(0.5)
(0.5)
= 56(.03125)( .125) = .00032
8 5
5
8-6
= 28(.015625)(. 25) = .00032
8C6(0.5) (0.5)
5
8-7
= 8(.0078125)(0.5) = .00032
8C7(0.5) (0.5)
5
8-8
= 1(.00390625)(1) = .00032
8C8(0.5) (0.5)
3.
x
6Cx
0
1
2
3
4
5
6
px
1
1
6
0.05
15
0.0025
20 0.000125
15 6.25E-06
6 3.13E-07
1 1.56E-08
q6-x
0.735092
0.773781
0.814506
0.857375
0.9025
0.95
1
x 6-x
6Cxp q
0.735091891
0.232134281
0.030543984
0.002143438
8.46094E-05
1.78125E-06
1.5625E-08
b)
E(x) = 6(0.05) = 0.3
5.
x
10Cx
0
1
2
3
4
5
6
7
8
9
10
px
1
1
10 0.083333
45 0.006944
120 0.000579
210 4.82E-05
252 4.02E-06
210 3.35E-07
120 2.79E-08
45 2.33E-09
10 1.94E-10
1 1.62E-11
q10-x
0.418904
0.456986
0.49853
0.543851
0.593292
0.647228
0.706067
0.770255
0.840278
0.916667
1
At least two 9’s, subtract zero 9’s and one 9’s from 1
1 – 0.418903888 – 0.380821716
= 0.2003
10Cxp
x 10-x
q
0.418903888
0.380821716
0.155790702
0.037767443
0.006008457
0.000655468
4.96567E-05
2.57957E-06
8.79398E-08
1.77656E-09
1.61506E-11
7.
a)
x
px
8Cx
0
1
2
3
4
5
6
7
8
1
8
28
56
70
56
28
8
1
1
0.65
0.4225
0.274625
0.178506
0.116029
0.075419
0.049022
0.031864
q8-x
0.000225
0.000643
0.001838
0.005252
0.015006
0.042875
0.1225
0.35
1
x 8-x
8Cxp q
0.000225188
0.003345643
0.021746682
0.080773392
0.187509659
0.278585779
0.258686795
0.137262381
0.031864481
At least four times, subtract 0 – 3 from one.
1 – 0.000225288 – 0.003345643 – 0.021746682 – 0.080773392
= 0.8939
b)
E(x) = 8(0.65) = 5.2
8.
a)
20C5(0.1)5(0.9)15
= 0.0319
b)
E(x) = 20(0.1)
=2
7.3 Geometric Distributions pg 394 # 1 - 11
Multliply the probability of the outcomes
Geometric because you are going until an event happens
P(x) = qxp
E(x) = q/p
1.
a) Yes, waiting for a 6
b) no, not going until an event
d) Yes e) no
2.
a)
Number of
unsuccessful
trials (x)
0
1
2
3
4
5
q^x
1
0.8
0.64
0.512
0.4096
0.32768
P
q^x
1
0.5
0.25
0.125
0.0625
0.03125
P
0.2
0.2
0.2
0.2
0.2
0.2
q^xp
0.2
0.16
0.128
0.1024
0.08192
0.065536
0.5
0.5
0.5
0.5
0.5
0.5
q^xp
0.5
0.25
0.125
0.0625
0.03125
0.015625
b)
Number of
unsuccessful
trials (x)
0
1
2
3
4
5
c) waiting, but no probability to use
3.
Number of
unsuccessful
trials (x)
0
1
2
q^x
1
0.916667
0.840278
P
0.083333
0.083333
0.083333
q^xp
0.083333
0.076389
0.070023
 3rd roll because 2 failures. This is the answer.
b)
E(x) = q/p = 11/12÷1/12 = 11/12 x 12/1 = 11/1 = 11
4.
a) Odds are 3:1, prob = ¾
(3/4)8 = .1001
b)
E(x) = q/p = 3/4÷1/4 = 3/4 x 4/1 = 3/1 = 3
5.
She’s expecting to roll a 2. The prob of rolling a 2 is 1/36, prob against is 35/36
E(x) = q/p = 35/36÷1/36 = 35/36 x 36/1 = 35/1 = 35
6.
Number of
unsuccessful
trials (x)
0
1
2
3
q^x
1
0.666667
0.444444
0.296296
P
0.333333
0.333333
0.333333
0.333333
q^xp
0.333333333
0.222222222
0.148148148
0.098765432
Prob of someone winning on the first 4 shows is 0.8025. (add up last row)
Prob of someone now winning is 1 – 0.8025 = 0.1975
7.
a)
Number of
unsuccessful
trials (x)
4
q^x
P
0.716393
0.08
q^xp
0.057311437
Use 4th failure for success on 5th roll
b)
Number of
unsuccessful
trials (x)
0
1
2
3
4
5
6
q^x
P
1
0.08
0.92
0.08
0.8464
0.08
0.778688
0.08
0.716393
0.08
0.659082
0.08
0.606355
0.08
q^xp
0.08
0.0736
0.067712
0.06229504
0.057311437
0.052726522
0.0485084
Prob of getting an A on the first 7 tests is 0.4422.
Prob of not getting an A is 1 – 0.4422 = 0.5578.
c) E(x) = .92 / .08 = 11.5 tests
8.
a)
Number of
unsuccessful
trials (x)
0
1
2
3
4
5
6
7
8
9
10
11
q^x
P
1
0.1
0.9
0.1
0.81
0.1
0.729
0.1
0.6561
0.1
0.59049
0.1
0.531441
0.1
0.478297
0.1
0.430467
0.1
0.38742
0.1
0.348678
0.1
0.313811
0.1
q^xp
0.1
0.09
0.081
0.0729
0.06561
0.059049
0.0531441
0.04782969
0.043046721
0.038742049
0.034867844
0.03138106
12 call means 11 misses. Therefore probability = 0.0314
b) E(x) = .9/.1 = 9 hang ups.
9.
Number of
unsuccessful
trials (x)
0
1
2
3
q^x
1
0.6
0.36
0.216
P
0.4
0.4
0.4
0.4
q^xp
0.4
0.24
0.144
0.0864
Prob of being on time first 0.8704
1 – 0.8704 = 0.1296
b)
E(x) = 0.6/0.4 = 1.5
On time on the 1.5th delivery
10.
a)
Number of
unsuccessful
trials (x)
0
1
2
3
4
q^x
P
1
0.34
0.66
0.34
0.4356
0.34
0.287496
0.34
0.189747
0.34
5 person means 4 misses. 0.0645
b)
E(x) = .66 / .34
= 1.9 people
q^xp
0.34
0.2244
0.148104
0.09774864
0.064514102
11.
a)
Number of
unsuccessful
trials (x)
3
q^x
P
0.941192
0.02
q^xp
0.01882384
q^x
P
1
0.02
0.98
0.02
0.9604
0.02
0.941192
0.02
q^xp
0.02
0.0196
0.019208
0.01882384
b)
Number of
unsuccessful
trials (x)
0
1
2
3
=0.02+0.0196+0.01921+0.01882
= 0.07763
c) E(x) = q/p = 49/50÷1/50 = 49/50 x 50/1 = 49/1 = 49
7.4 Hypergeometric Distributions Pg 404 #1 – 3, 5b, 6b, 7 – 11, 12ab
Uses dependent trials
𝑎𝐶𝑥
p(x) =
× 𝑛−𝑎𝐶𝑟−𝑥
𝑛𝐶𝑟
𝑟𝑎
E(X) = 𝑛
n is the total objects
r is your selection size
a is the size of the group you choose from
x is how many you select from the group in “a”
1.
a) yes, prob changes as cards dealt out.
b) no, rolls are independent
c) no, numbers are independent
d) no, screws are independent
e) yes, removing a name changes the probability
f) yes, removing a lefty changes the probability
g) yes, removing a lefty changes the probability
2.
a)
n
6
r
3
a
3
x
aCx
n-aCr-x
1
3
3
1
0
1
2
3
1
3
3
1
nCr
20
20
20
20
p(x)
0.05
0.45
0.45
0.05
nCr
56
56
56
56
p(x)
0.017857
0.267857
0.535714
0.178571
b)
n
8
r
3
a
5
x
aCx
n-aCr-x
1
3
3
1
0
1
2
3
1
5
10
10
3.
a)
n
12
r
4
x
2
aCx
10
a
5
n-aCr-x
21
nCr
p(x)
495 0.424242
b)
E(x) = ra/n
= 4(7)/12
= 2.3
5.
b)
n
60
x
r
8
aCx
5 1221759
6b)
E(x) = 50(200)/1200
= 8.3
a
45
n-aCr-x
455
nCb
2558620845
p(x)
0.217266
7.
a)
n
20
r
5
a
5
x
aCx
n-aCr-x
3003
1365
0
1
1
5
nCr
p(x)
15504 0.193692
15504 0.440209
=.1936+.4402
= 0.6338
b)
E(x) = ra/n
= 5(5)/20
= 1.25
8.
a)
n
16
r
5
x
2
a
5
aCx
10
n-aCr-x
165
nCr
p(x)
4368 0.377747
b)
E(x) = ra/n
= 5(5)/16
= 1.56
9.
a)
n
12
r
4
a
5
x
aCx
n-aCr-x
35
0
b)
E(x) = ra/n
= 5(5)/12
= 2.1
1
nCr
p(x)
495 0.070707
10.
a)
n
25
r
12
x
aCx
462
6
a
11
n-aCr-x
nCr
3003 5200300
p(x)
0.26679
b)
n
25
r
12
x
aCx
0
1
2
a
11
1
11
55
n-aCr-x
nCr
p(x)
91 5200300 0.0000175
364 5200300
0.00077
1001 5200300 0.010587
= 1 – 0.0000175 - .00077 – 0.0106
= 0.9886
c)
E(x) = ra/n
= 12(11)/25
= 5.3
11.
a)
n
52
r
7
x
aCx
3
a
13
286
n-aCr-x
82251
nCr
133784560
b)
E(x) = ra/n
= 7(13)/52
= 91/52
= 1.75
12
a)
n
11
r
4
a
5
x
aCx
n-aCr-x
1
4
5
nCr
p(x)
330 0.015152
b)
n
11
r
4
a
4
x
aCx
n-aCr-x
21
2
6
nCb
p(x)
330 0.381818
p(x)
0.175833
Review pg 406 #1 – 3, 5 – 10, 12 – 19, 21b
Practice Test pg 408 #1 – 3, 5 – 9