1) If we pick a student at random, P(S)=0.6 and P(O)=0.4
a) S and S and O has probability 0.6*0.6*0.4 = 0.144
b)
OOO
OOS
OSO
SOO
SSO
SOS
OSS
0.064
0.096
0.096
0.096
0.144
0.144
0.144
OOS for instance, is 0.4*0.4*0.6 etc.
SSS
0.216
c) X=0 is SSS
X=1 is “SSO or SOS or OSS” with probability 0.144+0.144+0.144
X=2 is “OOS or OSO or SOO” with probability 0.096+0.096+0.096
X=3 is OOO
X
0
1
2
3
.216
.432
.288
.064
d) X=2 or X=3 has probability 0.288+0.064 = 0.352
e)
𝜇𝑋 = 0 ∗ 0.216 + ⋯ + 3 ∗ 0.064 = 1.2
𝜇𝑋 2 = 02 ∗ 0.216 + ⋯ + 32 ∗ 0.064 = 2.16
𝜎𝑋 =
2.16 − 1.22 = 0.8485
2) Refer to problem 4.2. Pick four marbles at random. Write down the probability distribution for
X, the number of brown marbles. What are the mean and standard deviation for X?
2) b=brown (0.3), n=not brown (0.7)
nnnb (for instance) has probability 0.7*0.7*0.7*0.3
nnnb = 0.1029
bbbn = 0.0189
nnnn = 0.2401
bnnb = 0.0441
nnbn = 0.1029
bbnb =0.0189
bbbb = 0.0081
nbbn = 0.0441
nbnn = 0.1029
bnbb = 0.0189
bbnn = 0.0441
nbnb = 0.0441
bnnn = 0.1029
nbbb = 0.0189
nnbb = 0.0441
bnbn = 0.0441
X=0 is “nnnn”
X=1 is “nnnb or nnbn or nbnn or bnnn” = 0.1029 + 0.1029 + 0.1029 + 0.1029
X=2 is “bbnn or nnbb or bnnb or nbbn or nbnb or bnbn” = 0.0441 + 0.0441 + 0.0441 + 0.0441 +
0.0441 + 0.0441
X=3 is “bbbn or bbnb or bnbb or nbbb” = 0.0189 + 0.0189 + 0.0189 + 0.0189
X=4 is “bbbb”
X
0
1
2
3
4
.2401
.4116
.2646
.0756
.0081
𝜇𝑋 = 0 ∗ 0.2401 + ⋯ + 4 ∗ 0.0081 = 1.2
𝜇𝑋 2 = 02 ∗ 0.2401 + ⋯ + 42 ∗ 0.0081 = 2.28
2.28 − 1.22 = 0.9165
𝜎𝑋 =
3) (a) A=type A blood, N=not type A blood P(A) = 0.27 P(N) = 0.73
Going a little faster, X=0 is NNN with probability 0.73*0.73*0.73
X=3 is AAA with probability 0.27*0.27*0.27
X=1 is, for instance, NNA with probability 0.73*0.73*.27=0.143883
As we can see from the above examples, there will be 3 such calculations (the single A could be
in one of three slots), so we’ll be adding this answer to itself 3 times: 3*0.143883 = 0.431649
Likewise X=2 can come from AAN with probability 0.27*0.27+0.73 = 0.053217 which can
occur one of 3 ways (the single N can be in one of three slots), so 0.053217*3 = 0.159651
X
0
1
2
3
.389017
.431649
.159651
.019683
𝜇𝑋 = 0 ∗ .389017 + ⋯ + 3 ∗ .019683 = 0.81
𝜇𝑋 2 = 02 ∗ .389017 + ⋯ + 32 ∗ 019683 = 1.2474
𝜎𝑋 =
1.2474 − 0.812 = 0.7690
(b) A=type AB blood, N=not type AB blood P(A) = 0.2 P(N) = 0.8
X=0 is NNN with probability 0.8*0.8*0.8*.8
X=4 is AAA with probability 0.2*0.2*0.2*.2
X=1 is, for instance, NNNA with probability 0.8*0.8*.8*.2=0.1024
As we can see from the above examples, there will be 4 such calculations (the single A could be
in one of fur slots), so we’ll be adding this answer to itself 4 times: 4*0.1024 = .4096
Likewise X=3 can come from AAAN with probability 0.2*0.2*.2*.8 = 0.0064 which can occur
one of 4 ways (the single N can be in one of four slots), so 0.0064*4 = 0.0256
X=2 can be worked out by the process of elimination (rather than listing out all six cases; eg,
AANN) since they must add up to 1. Subtract the other four cases from 1 to get the middle case.
Y
0
1
2
3
4
.4096
.4096
.1536
.0256
.0016
𝜇𝑋 = 0 ∗ .4096 + ⋯ + 4 ∗ .0016 = 0.8
𝜇𝑋 2 = 02 ∗ .4096 + ⋯ + 42 ∗ .0016 = 1.28
𝜎𝑋 =
1.2474 − 0.82 = 0.8
4) For the coin, P(“1”) = ½ and P(“2”)= ½
For the die, the probability of any particular number 1…6 is 1/6.
sum 1
2
3
4
5
6
2
3
4
5
6
7
1
3
4
5
6
7
8
2
Each of these has probability 1/2 * 1/6=1/12. The sum of 2 can only occur one way; likewise the
sum of 8. For any sum of 3,4,5,6,or 7 it can occur two ways (eg, 5 can be 1+4 or 2+3), so its
probability will be 1/12 + 1/12 = 2/12 = 1/6:
2
3
1/12 1/6
4
1/6
5
1/6
6
1/6
7
1/6
8
1/12
b)
𝜇𝑋 = 2 ∗ 1/12 + ⋯ + 8 ∗ 1/12 = 5
𝜇𝑋 2 = 22 ∗ 1/12 + ⋯ + 82 ∗ 1/12 = 28.1666667
𝜎𝑋 =
28.1666667 − 52 = 1.7795
c) The mean is the long-term average of many repeats, so 750 rolls * 5 number/roll = 3750
5) The probability of a boy (B) is 0.5, as is a girl (G). Possibilities are:
G (0.5) or BG (0.5*0.5=0.25) or BBG (0.5*0.5*0.5=0.125) or BBB (0.5*0.5*0.5=0.125)
X = number of children (1 or 2 or 3, which is BBG or BBB, add probabilities):
X
1
2
3
0.5
0.25
0.250
Y = number of boys (0 or 1 or 2 or 3, from G or BG or BBG or BBB respectively):
Y
0
1
2
3
0.5
0.25
0.125
.125
𝜇𝑋 = 1 ∗ 0.5 + 2 ∗ 0.25 + 3 ∗ 0.25 = 1.75
𝜇𝑋 2 = 12 ∗ 0.5 + 22 ∗ 0.25 + 32 ∗ 0.25 = 3.75
𝜎𝑋 =
3.75 − 1.752 = 0.8292
1.75 is the average of many “repeats of picking an X”, so:
400 couples * (about, for large n) 1.75 children/couple = 700 children
6) X = winnings ($)
We must break into non-overlapping cases:
All 52 cards are equally likely. Half (26) are red (X=0). Of the black cards, a quarter (13) are
spades (X=5), and 13 are clubs. 12 clubs are non-aces (10) and 1 card is the ace of clubs (10 +
20 total $).
X
prob.
0
26/52 = 1/2
5
13/52 = 1/4
10
12/52
30
1/52
1
𝜇𝑋 = 0 ∗ + ⋯ + 30 ∗ 1/52 = 4.134615
2
1
𝜇𝑋 2 = 02 ∗ + ⋯ + 302 ∗ 1/52 = 46.634615
2
𝜎𝑋 =
46.634615 − 4.134612 = 5.43503
1.75 is the average of many “repeats of picking an X”, so:
300 games * (about, for large n) 4.134615 dollars/game = $1240.38
If the person running this game wants to make (in the long run) about $1 per game, he should
charge:
$___ to play each game – $4.13 paid back per game (in the long run) = $1 profit, so $5.13
7) 0.2 vs. 0.8 is W vs. L for race one, and 0.3 vs. 0.7 is W vs. L for race two:
WW
0.2*0.3 = 0.06
WL
0.2 * 0.7 = 0.14
LW
0.8 * 0.3 = 0.24
LL
0.8*0.7 = 0.56
X = value of horse:
win one race = WL or LW, add probabilities:
X
100,000
50,000
10,000
.06
.38
.56
Y
80,000
.06
30,000
.38
–10,000
.56
Notice that Y is (in any case) X – 20,000
𝜇𝑋 = 100,000 ∗ 0.06 + ⋯ + 10,000 ∗ 0.56 = 30,600
𝜇𝑋 2 = 100,0002 ∗ 0.06 + ⋯ + 10,0002 ∗ 0.56 = 1606000000
𝜎𝑋 =
1606000000 − 306002 = 25,977.40
Alternately, working in “thousands of dollars” will bring these numbers down to size.
The mean of Y is the mean of X minus 20,000 (as you can see by computing).
The standard deviation of Y is the same as the standard deviation of X (as you can see by
computing; we will study this in the next section, and as we did for data in section 2; their values
are equally spread out, just shifted).
8)
X
0
0.75
3
0.25
𝜇𝑋 = 0 ∗ 0.75 + 3 ∗ 0.25 = 0.75
𝜇𝑋 2 = 02 ∗ 0.75 + 32 ∗ 0.25 = 2.25
𝜎𝑋 =
2.25 − 0.752 = 1.29904
The casino takes in $1 per game and pays back $0.75 per game (in the long run), netting 1 – 0.75
= 0.25 per game. The longer they let this go on (the casino plays very many games), the truer
this will be.