Section 8.4
For this section there are basically THREE key equations; and this is based
n (which is the number of trials or times the binomial experiment will be
accomplished), on p (which is the probability of “success”, though success
on
in this case means that which you are seeking an answer, not necessarily
q
success on an experiment), and on (which is the probability of “failure”,
or the opposite of “success”…it is equal to (1 – p):
The first calculation is for calculating an exact number of successes (which is
designated by “k”):
P(k) = C(n,k) * (p)k * (q)(n-k)
The second equation is for calculating the mean (µ) for the binomial
experiment:
µ=n*p
And the third equation is for calculating the standard deviation (σ) for the
binomial experiment:
σ = sqrt(n * p * q)
Problems 76-85 are simply calculating the probabilities of exactly 0 and 1
using the first equation above…
Problem #76
P(0) = C(5,0)*(0.4)0*(0.6)5 = 0.0777
P(1) = C(5,1)*(0.4)1*(0.6)4 = 0.2592
Problem #77
P(0) = C(6,0)*(0.2)0*(0.8)6 = 0.2621
P(1) = C(6,1)*(0.2)1*(0.8)5 = 0.3932
Problem #78
P(0) = C(8,0)*(0.5)0*(0.5)8 = 0.0039
P(1) = C(8,1)*(0.5)1*(0.5)7 = 0.0312
Problem #79
P(0) = C(10,0)*(0.3)0*(0.7)10 = 0.0282
P(1) = C(10,1)*(0.3)1*(0.7)9 = 0.1211
Problem #80
P(0) = C(4,0)*(0.1)0*(0.9)4 = 0.6561
P(1) = C(4,1)*(0.1)1*(0.9)3 = 0.2916
Problem #81
P(0) = C(8,0)*(0.2)0*(0.8)8 = 0.1678
P(1) = C(8,1)*(0.2)1*(0.8)7 = 0.3355
Problem #82
P(0) = C(3,0)*(0.8)0*(0.2)3 = 0.0080
P(1) = C(3,1)*(0.8)1*(0.2)2 = 0.0960
Problem #83
P(0) = C(7,0)*(0.9)0*(0.1)7 = 0.0000001
P(1) = C(7,1)*(0.9)1*(0.1)6 = 0.0000063
Problem #84
P(0) = C(12,0)*(0.8)0*(0.2)12 = 0.000000004
P(1) = C(12,1)*(0.8)1*(0.2)11 = 0.000000196
Problem #85
P(0) = C(2,0)*(0.6)0*(0.4)2 = 0.1600
P(1) = C(2,1)*(0.6)1*(0.4)1 = 0.4800
Problems 86-90 are problems that want “at least one”, which means they
want all the probabilities calculated from 1 to n (or in other words, they want
all the probabilities calculated except P(0). So to simplify the work, we will
use the complement rule first (1 – P(0) ), but show that by calculating all the
rest using that first equation will give you the same answer…
Problem #86
P(at least 1) = 1 – C(8,0)*(0.5)0*(0.5)8 = 1 – 0.0039 = 0.9961
or
P(1) = C(8,1)*(0.5)1*(0.5)7 = 0.0312
P(2) = C(8,2)*(0.5)2*(0.5)6 = 0.1094
P(3) = C(8,3)*(0.5)3*(0.5)5 = 0.2188
P(4) = C(8,4)*(0.5)4*(0.5)4 = 0.2734
P(5) = C(8,5)*(0.5)5*(0.5)3 = 0.2188
P(6) = C(8,6)*(0.5)6*(0.5)2 = 0.1094
P(7) = C(8,7)*(0.5)7*(0.5)1 = 0.0312
P(8) = C(8,8)*(0.5)8*(0.5)0 = 0.0039
P(at least 1) = 0.0312 + 0.1094 + 0.2188 + 0.2734 + 0.2188 + 0.1094 +
0.0312 + 0.0039 = 0.9961
Problem #87
P(at least 1) = 1 – C(3,0)*(0.8)0*(0.2)3 = 1 – 0.0080 = 0.9920
or
P(1) = C(3,1)*(0.8)1*(0.2)2 = 0.0960
P(2) = C(3,2)*(0.8)2*(0.2)1 = 0.3840
P(3) = C(3,3)*(0.8)3*(0.2)0 = 0.5120
P(at least 1) = 0.0960 + 0.3840 + 0.5120 = 0.9920
Problem #88
P(at least 1) = 1 – C(12,0)*(0.8)0*(0.2)12 = 1 – 0.0000000041 = 0.9999999959
≈1
or
P(1) = C(12,1)*(0.8)1*(0.2)11 = 0.0000002
P(2) = C(12,2)*(0.8)2*(0.2)10 = 0.0000043
P(3) = C(12,3)*(0.8)3*(0.2)9 = 0.0000577
P(4) = C(12,4)*(0.8)4*(0.2)8 = 0.0005190
P(5) = C(12,5)*(0.8)5*(0.2)7 = 0.0033219
P(6) = C(12,6)*(0.8)6*(0.2)6 = 0.0155021
P(7) = C(12,7)*(0.8)7*(0.2)5 = 0.0531502
P(8) = C(12,8)*(0.8)8*(0.2)4 = 0.1328756
P(9) = C(12,8)*(0.8)9*(0.2)3 = 0.2362232
P(10) = C(12,8)*(0.8)10*(0.2)2 = 0.2834678
P(11) = C(12,8)*(0.8)11*(0.2)1 = 0.2061584
P(12) = C(12,8)*(0.8)12*(0.2)0 = 0.0687195
P(at least 1) = 0.0000002 + 0.0000043 + 0.0000577 + 0.0005190 +
0.0033219 + 0.0155021 + 0.0531502 + 0.1328756 + 0.2362232 +
0.2834678 + 0.2061584 + 0.0687195 = 0. 9999999959 ≈ 1
Problem #89
P(at least 1) = 1 – C(7,0)*(0.9)0*(0.1)7 = 1 – 0.0000001 = 0.9999991
or
P(1) = C(7,1)*(0.9)1*(0.1)6 = 0.0000063
P(2) = C(7,2)*(0.9)2*(0.1)5 = 0.0001701
P(3) = C(7,3)*(0.9)3*(0.1)4 = 0.0025515
P(4) = C(7,4)*(0.9)4*(0.1)3 = 0.0229635
P(5) = C(7,5)*(0.9)5*(0.1)2 = 0.1240029
P(6) = C(7,6)*(0.9)6*(0.1)1 = 0.3720087
P(7) = C(7,7)*(0.9)7*(0.1)0 = 0.4782969
P(at least 1) = 0.0000063 + 0.0001701 + 0.0025515 + 0.0229635 +
0.1240029 + 0.3720087 + 0.4782969 = 0. 9999991
Problem #90
P(at least 1) = 1 – C(5,0)*(0.4)0*(0.6)5 = 1 – 0.0778 = 0.9222
or
P(1) = C(5,1)*(0.4)1*(0.6)4 = 0.2592
P(2) = C(5,2)*(0.4)2*(0.6)3 = 0.3456
P(3) = C(5,3)*(0.4)3*(0.6)2 = 0.2304
P(4) = C(5,4)*(0.4)4*(0.6)1 = 0.0768
P(5) = C(5,5)*(0.4)5*(0.6)0 = 0.0102
P(at least 1) = 0.2592 + 0.3456 + 0.2304 + 0.0768 + 0.0102 = 0.9222
Problems 91-95 are problems that want a “probability distribution”, which
means they want all the probabilities calculated from 0 to n. And then place
it in a “table”…
Problem #91
P(0) = C(5,0)*(0.4)0*(0.6)5 = 0.0778
P(1) = C(5,1)*(0.4)1*(0.6)4 = 0.2592
P(2) = C(5,2)*(0.4)2*(0.6)3 = 0.3456
P(3) = C(5,3)*(0.4)3*(0.6)2 = 0.2304
P(4) = C(5,4)*(0.4)4*(0.6)1 = 0.0768
P(5) = C(5,5)*(0.4)5*(0.6)0 = 0.2188
X
P(X)
0
1
2
3
4
5
.0778 .2592 .3456 .2304 .0768 .2188
Problem #92
P(0) = C(3,0)*(0.2)0*(0.8)8 = 0.512
P(1) = C(3,1)*(0.2)1*(0.8)7 = 0.384
P(2) = C(3,2)*(0.2)2*(0.8)6 = 0.096
P(3) = C(3,3)*(0.2)3*(0.8)5 = 0.008
X
P(X)
0
.512
1
.384
2
.096
3
.008
Problem #93
P(0) = C(8,0)*(0.2)0*(0.8)8 = 0.1678
P(1) = C(8,1)*(0.2)1*(0.8)7 = 0.3355
P(2) = C(8,2)*(0.2)2*(0.8)6 = 0.2936
P(3) = C(8,3)*(0.2)3*(0.8)5 = 0.1468
P(4) = C(8,4)*(0.2)4*(0.8)4 = 0.0459
P(5) = C(8,5)*(0.2)5*(0.8)3 = 0.0092
P(6) = C(8,6)*(0.2)6*(0.8)2 = 0.0011
P(7) = C(8,7)*(0.2)7*(0.8)1 = 0.00008
P(8) = C(8,8)*(0.2)8*(0.8)0 = 0.000003
X
0
1
2
3
4
5
6
7
8
P(X) .1678 .3355 .2936 .1468 .0459 .0092 .0011 .00008 .000003
Problem #94
P(0) = C(8,0)*(0.8)0*(0.2)8 = 0.000003
P(1) = C(8,1)*(0.8)1*(0.2)7 = 0.00008
P(2) = C(8,2)*(0.8)2*(0.2)6 = 0.0011
P(3) = C(8,3)*(0.8)3*(0.2)5 = 0.0092
P(4) = C(8,4)*(0.8)4*(0.2)4 = 0.0459
P(5) = C(8,5)*(0.8)5*(0.2)3 = 0.1468
P(6) = C(8,6)*(0.8)6*(0.2)2 = 0.2936
P(7) = C(8,7)*(0.8)7*(0.2)1 = 0.3355
P(8) = C(8,8)*(0.8)8*(0.2)0 = 0.1678
X
0
1
2
3
4
5
6
7
8
P(X) .000003 .00008 .0011 .0092 .0459 .1468 .2936 .3355 .1678
Problem #95
P(0) = C(3,0)*(0.8)0*(0.2)8 = 0.008
P(1) = C(3,1)*(0.8)1*(0.2)7 = 0.096
P(2) = C(3,2)*(0.8)2*(0.2)6 = 0.384
P(3) = C(3,3)*(0.8)3*(0.2)5 = 0.512
X
P(X)
0
.008
1
.096
2
.384
3
.512
Problems 96-115 are problems that are designed to use table 1 in the back of
the book. The table is designed such that when you choose a number for “k”
on the table, it gives you the sum of all the probabilities from zero to that
number.
So for example, if you have n=6 and p=0.4 and you want all the values from
zero to 4 (P<4), then you go to the part of the table that says “n = 6 trials”;
and within that box we will only use the column under 0.4… now since we
want the sum of all the values from 0 to 4 we look up a k of 4, and you see a
value of .959…which is equal to P(0)+P(1)+P(2)+P(3)+P(4)…simpler than
calculating each value as we did above and adding them together.
To calculate some of the problems below there will have to be some
manipulations…
For Problems 96-105, n=10 and p=0.4
Problem #96
P[Y<6] means that I want all the probabilities from 0 to 6, which means I
choose a k of 6, giving an answer of: 0.945
Problem #97
P[Y<6] means that I want all the probabilities from 0 to 5, which means I
choose a k of 5, giving an answer of: 0.834
Problem #98
P[Y<8] means that I want all the probabilities from 0 to 8, which means I
choose a k of 8, giving an answer of: 0.998
Problem #99
P[Y>5] means that I want all the probabilities from 5 to 10, which is NOT
zero to a number. But we know that from 0 to 10 equals a probability of
ONE, because all the probabilities must add up to one. So is we get rid of
the part of that that we do not want (which in this case is 0 to 4) we would
have the answer. So if we look up k equal to 4 (which will give us the
probability of 0 to 4) we get: means I choose a k of 6, giving an answer of:
0.633; so the answer to this problem is 1 – 0.633 = 0.267
Problem #100
P[Y>3] means that I want all the probabilities from 4 to 10, which is NOT
zero to a number. But we know that from 0 to 10 equals a probability of
ONE, because all the probabilities must add up to one. So is we get rid of
the part of that that we do not want (which in this case is 0 to 3) we would
have the answer. So if we look up k equal to 3 (which will give us the
probability of 0 to 3) we get: means I choose a k of 3, giving an answer of:
0.382; so the answer to this problem is 1 – 0.382 = 0.618
Problem #101
P[Y=7] means that I want all the probability of 7 (nothing else), which is
NOT zero to a number. We could just calculate it like we did in the previous
exercises, but we need to realize that using the table is easier…we know that
if we choose k equal to 7 that it is the same as
P(0)+P(1)+P(2)+P(3)+P(4)+P(5)+P(6)+P(7), and that if k equals 6 it is:
P(0)+P(1)+P(2)+P(3)+P(4)+P(5)+P(6)…so if you subtract these you get:
P(7)…the answer…so for k=7 it is 0.988 and for k=6 it is 0.945; so the
answer to this problem is 0.988 – 0.945 = 0.043
(If you calculated using the equation you get 0.042467…the slight bit off is
rounding)
Problem #102
P[Y=9] means that I want all the probability of 9 (nothing else), which is
NOT zero to a number. We could just calculate it like we did in the previous
exercises, but we need to realize that using the table is easier…we know that
if we choose k equal to 9 that it is the same as
P(0)+P(1)+P(2)+P(3)+P(4)+P(5)+P(6)+P(7)+P(8)+P(9),
and that if k equals 8 it is:
P(0)+P(1)+P(2)+P(3)+P(4)+P(5)+P(6)+P(7)+P(8)
…so if you subtract these you get:
P(9)…the answer…so for k=9 it is 1 (“1-“ means when rounded it is close to
one but not quite 1) and for k=8 it is 0.998; so the answer to this problem is
1 – 0.998 = 0.002
(If you calculated using the equation you get 0.001573)
Problem #103
P[4<Y<6] means that I want all the probability of 4 to 6, which is NOT zero
to a number. We know that if we choose k equal to 6 that it is the same as
P(0)+P(1)+P(2)+P(3)+P(4)+P(5)+P(6), which has too much, it also has
P(0)+P(1)+P(2)+P(3), which is the same as if you chose a k equal to 3
…so if you subtract these you get the answer to this problem is
0.945 – 0.382 = 0.563
Problem #104
P[Y>9] means that I want all the probability of 10, which is NOT zero to a
number. But we know that from 0 to 10 equals a probability of ONE,
because all the probabilities must add up to one. So is we get rid of the part
of that that we do not want (which in this case is 0 to 9) we would have the
answer. So if we look up k equal to 9 (which will give us the probability of
0 to 9) we get: means I choose a k of 9, giving an answer of: “1-” or 1; so the
answer to this problem is 1 - 1 = 0
Problem #105
P[Y<9] means that I want all the probabilities from 0 to 8, which means I
choose a k of 8, giving an answer of: 0.998
For Problems 106-115, n=15 and p=0.6…remember that the table is split
into two pieces (unfortunately)
Problem #106
P[Y<9] means that I want all the probabilities from 0 to 9, which means I
choose a k of 9, giving an answer of: 0.597
Problem #107
P[Y>8] means that I want all the probabilities from 9 to 15, which is NOT
zero to a number. But we know that from 0 to 15 equals a probability of
ONE, because all the probabilities must add up to one. So is we get rid of
the part of that that we do not want (which in this case is 0 to 8) we would
have the answer. So if we look up k equal to 8 (which will give us the
probability of 0 to 8) we get: means I choose a k of 8, giving an answer of:
0.390; so the answer to this problem is 1 – 0.390 = 0.610
Problem #108
P[Y<12] means that I want all the probabilities from 0 to 12, which means I
choose a k of 12, giving an answer of: 0.973
Problem #109
P[Y>10] means that I want all the probabilities from 10 to 15, which is NOT
zero to a number. But we know that from 0 to 15 equals a probability of
ONE, because all the probabilities must add up to one. So is we get rid of
the part of that that we do not want (which in this case is 0 to 9) we would
have the answer. So if we look up k equal to 9 (which will give us the
probability of 0 to 9) we get: means I choose a k of 9, giving an answer of:
0.597; so the answer to this problem is 1 – 0.597 = 0.403
Problem #110
P[Y>13] means that I want all the probabilities from 14 to 15, which is NOT
zero to a number. But we know that from 0 to 15 equals a probability of
ONE, because all the probabilities must add up to one. So is we get rid of
the part of that that we do not want (which in this case is 0 to 13) we would
have the answer. So if we look up k equal to 13 (which will give us the
probability of 0 to 13) we get: means I choose a k of 13, giving an answer
of: 0.995; so the answer to this problem is 1 – 0.995 = 0.005
Problem #111
P[9<Y<12] means that I want all the probability of 10 to 12, which is NOT
zero to a number. We know that if we choose k equal to 12 that it is the
same as
P(0)+P(1)+P(2)+P(3)+P(4)+P(5)+P(6)+P(7)+P(8)+P(9)+P(10)+P(11)+P(12),
which has too much, it also has
P(0)+P(1)+P(2)+P(3)+P(4)+P(5)+P(6)+P(7)+P(8)+P(9), which is the same as
if you chose a k equal to 9…so if you subtract these you get the answer to
this problem is 0.973 – 0.597 = 0.376
Problem #112
P[Y<9] means that I want all the probabilities from 0 to 8, which means I
choose a k of 8, giving an answer of: 0.390
Problem #113
P[8<Y<12] means that I want all the probability of 8 to 12, which is NOT
zero to a number. We know that if we choose k equal to 12 that it is the
same as
P(0)+P(1)+P(2)+P(3)+P(4)+P(5)+P(6)+P(7)+P(8)+P(9)+P(10)+P(11)+P(12),
which has too much, it also has
P(0)+P(1)+P(2)+P(3)+P(4)+P(5)+P(6)+P(7), which is the same as if you
chose a k equal to 7…so if you subtract these you get the answer to this
problem is 0.973 – 0.213 = 0.760
Problem #114
P[Y>9] means that I want all the probabilities from 10 to 15, which is NOT
zero to a number. But we know that from 0 to 15 equals a probability of
ONE, because all the probabilities must add up to one. So is we get rid of
the part of that that we do not want (which in this case is 0 to 9) we would
have the answer. So if we look up k equal to 9 (which will give us the
probability of 0 to 9) we get: means I choose a k of 9, giving an answer of:
0.597; so the answer to this problem is 1 – 0.597 = 0.403
Problem #115
P[Y<7] means that I want all the probabilities from 0 to 6, which means I
choose a k of 6, giving an answer of: 0.095
Problems #116-120
For these problems you will use two equations…
for the mean it is:  = n*p
and for standard deviation it is:  = √𝑛 ∗ 𝑝 ∗ 𝑞
(remembering that q is 1-p). So now to the problems…
Problem #116
(mean)  = n*p = 20*0.2 = 4
(standard deviation)  = √20 ∗ 0.2 ∗ 0.8 = 1.7889
Problem #117
(mean)  = n*p = 50*0.6 = 30
(standard deviation)  = √50 ∗ 0.6 ∗ 0.4 = 3.4641
Problem #118
(mean)  = n*p = 25*0.4 = 10
(standard deviation)  = √25 ∗ 0.4 ∗ 0.6 = 2.4495
Problem #119
(mean)  = n*p = 15*0.4 = 6
(standard deviation)  = √15 ∗ 0.4 ∗ 0.6 = 1.8974
Problem #120
(mean)  = n*p = 8*0.5 = 4
(standard deviation)  = √8 ∗ 0.5 ∗ 0.5 = 1.4142
Problem #121
The following is the tree for this problem (I used Heads and Tails to
represent the binomial choices, since flipping a coin is a binomial
experiment, and I assumed P[Heads] = 0.8)…
H
T
H
H
H
T
T
H
H
H
0.8
T
T
H
T
T
H
H
0.2
0.8
H
H
T
H
T
T
T
H
T
H
T
H
T
H
H
H
0.2
0.8
H
T
0.2
T
T
H
T
T
T
H
H
T
T
H
T
H
H
H
T
H
T
T
H
T
H
T
H
T
T
T
Problems #122-126
Since it says “…the next 15 bills…” then n=15; and since it says “…that
10% contain overcharges…” then p=0.10; and this is what will be used with
the appropriate part of table 1 to solve the problems…
Problem #122
“…at most three bills with overcharges…” is the same as saying P[Y<3]. So
that means the probabilities of 0 to 3, which means I choose a k of 3, giving
an answer of: 0.944
Problem #123
“…at least two bills with overcharges…” is the same as saying P[Y>2]. So
that means the probabilities of 2 to 15, which is NOT zero to a number. But
we know that from 0 to 15 equals a probability of ONE, because all the
probabilities must add up to one. So is we get rid of the part of that that we
do not want (which in this case is 0 to 1) we would have the answer. So if
we look up k equal to 1 (which will give us the probability of 0 to 1) we get:
means I choose a k of 1, giving an answer of: 0.549; so the answer to this
problem is 1 – 0.549 = 0.451
Problem #124
“…at least one bill with an overcharge, but no more than four bills…” is the
same as saying P[1<Y<4]. So that means the probabilities of 1 to 4, which is
NOT zero to a number. We know that if we choose k equal to 4 that it is the
same as :P(0)+P(1)+P(2)+P(3)+P(4), which has too much, it also has
P(0), which is the same as if you chose a k equal to 0…so if you subtract
these you get the answer to this problem is 0.987 – 0.206 = 0.781
Problem #125
(mean)  = n*p = 15*0.1 = 1.5
(standard deviation)  = √15 ∗ 0.1 ∗ 0.9 = 1.1619
Problem #126
Since the mean is 1.5 (from problem #125) and the standard deviation is
1.1619 (also from problem #125), then 2 standard deviations is simply:
2*1.1619 = 2.3238. And to go that far from the mean says that I am to add
and subtract that from the mean:
1.5 – 2.3238 = -0.7762 and 1.5 + 2.3238 = 3.8238
Since the probability distribution does not start until 0 and the highest value
does not reach 4 then it looks like P[Y<4] or P[Y<3], which means the
probabilities of 0 to 3, which is the same as the answer to problem# 12,
which is 0.944 which is (as Chebychev says) “at least 0.75”.
Problems #127-130
Since it says “…the next ten such putts…” then n=10; and since it says
“…make 83% of their three-foot putts…” then p=0.83; for this we cannot
use table 1 to solve the problems, so we are left we calculating using the
equation we used in the first few problems…
Problem #127
“…at least seven…” is the same as saying P[Y>7]. So that means the
probabilities of 7 to 10:
P(7) = C(10,7)*(0.83)7*(0.17)3 = 0.1600
P(8) = C(10,8)*(0.83)8*(0.17)2 = 0.2929
P(9) = C(10,9)*(0.83)9*(0.17)1 = 0.3178
P(10) = C(10,10)*(0.83)10*(0.17)0 = 0.1552
which means the answer is: 0.1600+0.2929+0.3178+0.1552 = 0.9259
Problem #128
“…at most eight…” is the same as saying P[Y<8]. So that means the
probabilities of 0 to 8, or it means everything except 9 and 10:
P(0) = C(10,0)*(0.83)0*(0.17)10 = 0.00000002
P(1) = C(10,1)*(0.83)1*(0.17)9 = 0.000001
P(2) = C(10,2)*(0.83)2*(0.17)8 = 0.00002
P(3) = C(10,3)*(0.83)3*(0.17)7 = 0.0003
P(4) = C(10,4)*(0.83)4*(0.17)6 = 0.0024
P(5) = C(10,5)*(0.83)5*(0.17)5 = 0.0141
P(6) = C(10,6)*(0.83)6*(0.17)4 = 0.0573
P(7) = C(10,7)*(0.83)7*(0.17)3 = 0.1600
P(8) = C(10,8)*(0.83)8*(0.17)2 = 0.2929
which means the answer is:
0.00000002+0.000001+0.00002+0.0003+0.0024+
0.0141+0.0573+0.1600+0.2929 = 0.5270
OR
P(9) = C(10,9)*(0.83)9*(0.17)1 = 0.3178
P(10) = C(10,10)*(0.83)10*(0.17)0 = 0.1552
which means the answer is: 1-(0.3178+0.1552) = 0.5270
Problem #129
“…at least five…” is the same as saying P[Y<5]. So that means the
probabilities of 0 to 5, or it means everything except 6 to 10:
P(0) = C(10,0)*(0.83)0*(0.17)10 = 0.00000002
P(1) = C(10,1)*(0.83)1*(0.17)9 = 0.000001
P(2) = C(10,2)*(0.83)2*(0.17)8 = 0.00002
P(3) = C(10,3)*(0.83)3*(0.17)7 = 0.0003
P(4) = C(10,4)*(0.83)4*(0.17)6 = 0.0024
P(5) = C(10,5)*(0.83)5*(0.17)5 = 0.0141
which means the answer is:
0.00000002+0.000001+0.00002+0.0003+0.0024+0.0141= 0.0168
OR
P(6) = C(10,6)*(0.83)6*(0.17)4 = 0.0573
P(7) = C(10,7)*(0.83)7*(0.17)3 = 0.1600
P(8) = C(10,8)*(0.83)8*(0.17)2 = 0.2929
P(9) = C(10,9)*(0.83)9*(0.17)1 = 0.3178
P(10) = C(10,10)*(0.83)10*(0.17)0 = 0.1552
which means the answer is:
1-(0.0573+0.1600+0.2929+0.3178+0.1552) = 0.0168
Problem #130
“…at least six but no more than eight…” is the same as saying P[6<Y<8].
So that means the probabilities of 6 to 8:
P(6) = C(10,6)*(0.83)6*(0.17)4 = 0.0573
P(7) = C(10,7)*(0.83)7*(0.17)3 = 0.1600
P(8) = C(10,8)*(0.83)8*(0.17)2 = 0.2929
which means the answer is:
0.0573+0.1600+0.2929 = 0.5102
Problems #134-137
Since it says “…sold ten of these…” then n=10; and since it says “…Twenty
percent…need warranty work…” then p=0.20; and this is what will be used
with the appropriate part of table 1 to solve the problems…
Problem #134
“…at most five need warranty work…” is the same as saying P[Y<5]. So
that means the probabilities of 0 to 5, which means I choose a k of 5, giving
an answer of: 0.953
Problem #135
“…at least two will need warranty work…” is the same as saying P[Y>2].
So that means the probabilities of 2 to 10, which is NOT zero to a number.
But we know that from 0 to 10 equals a probability of ONE, because all the
probabilities must add up to one. So is we get rid of the part of that that we
do not want (which in this case is 0 to 1) we would have the answer. So if
we look up k equal to 1 (which will give us the probability of 0 to 1) we get:
means I choose a k of 1, giving an answer of: 0.376; so the answer to this
problem is 1 – 0.376 = 0.624
Problem #136
The “expected number” is the same as the mean, which is:
 = n*p = 10*0.2 = 2
Problem #137
(standard deviation)  = √10 ∗ 0.2 ∗ 0.8 = 1.2649
Problems #138
Since it says “…13 of 20…” then n=20; and since it says “…80%
chance…without a cold…” then p=0.80; and this is what will be used with
the appropriate part of table 1 to solve the problem…
“…at least 13…” is the same as saying P[Y>13]. So that means the
probabilities of 13 to 20, which is NOT zero to a number. But we know that
from 0 to 10 equals a probability of ONE, because all the probabilities must
add up to one. So is we get rid of the part of that that we do not want (which
in this case is 0 to 12) we would have the answer. So if we look up k equal
to 1 (which will give us the probability of 0 to 12) we get: means I choose a
k of 12, giving an answer of: 0.032; so the answer to this problem is
1 – 0.032 = 0.968
Problems #139(a)
This requires the use of combinations (since order does not matter)…so
when choosing any 5 from the 50 you get all the possible outcomes, which
is: C(50,5) = 2118760
And to choose “2 or more defective” is to choose 2 defective OR 3 defective
OR 3 defective OR 5 defective…or the compliment to that is to choose 0
defective or 1 defective and subtract that from the whole (and the probability
in either case is this number divided by the total):
2 defective: C(12,2)*C(38,3) = 556776
3 defective: C(12,3)*C(38,2) = 154660
4 defective: C(12,4)*C(38,1) = 18810
5 defective: C(12,5)*C(38,0) = 792
Answer: (556776+154660+18810+792)/2118760 = 0.3450
OR
0 defective: C(12,0)*C(38,5) = 501942
1 defective: C(12,1)*C(38,4) = 885780
Answer: (1-(501942+885780))/2118760 = 0.3450
Problems #139(b)
Since it says “…selecting five items…” then n=5; and since it says “…box
of 50 has 123 defective…” then p= 12/50 = 0.24; and this is will require the
use of the equation…
And to choose “2 or more defective” is to choose 2 defective OR 3 defective
OR 3 defective OR 5 defective…or the compliment to that is to choose 0
defective or 1 defective and subtract that from 1:
2 defective: P(2) = C(5,2)*0.242*0.763 = 0.2529
3 defective: P(3) = C(5,3)*0.243*0.762 = 0.0798
4 defective: P(4) = C(5,4)*0.244*0.761 = 0.0126
5 defective: P(5) = C(5,5)*0.245*0.760 = 0.0008
Answer: 0.2529+0.0798+0.0126+0.0008 = 0.3461
OR
0 defective: P(0) = C(5,0)*0.240*0.765 = 0.2536
1 defective: P(1) = C(5,1)*0.241*0.764 = 0.4003
Answer: 1-(0.2536+0.4003) = 0.3461
Problems #140-143
Since it says “…15 homes…were selected…” then n=15; and since it says
“…10%…have unsafe levels…” then p=0.10; and this is what will be used
with the appropriate part of table 1 to solve the problems…
Problem #140
“…exactly two…” means P[Y=2] means that I want all the probability of 2
(nothing else), which is NOT zero to a number. We could just calculate it
like we did in the previous exercises, but we need to realize that using the
table is easier…we know that if we choose k equal to 2 that it is the same as
P(0)+P(1)+P(2),
and that if k equals 1 it is:
P(0)+P(1)
…so if you subtract these you get:
P(2)…the answer…so for k=2 it is 0.816 and for k=1 it is 0.549; so the
answer to this problem is 0.816 – 0.549 = 0.267
(If you calculated using the equation you get 0.2669)
Problem #141
“…at least one…” is the same as saying P[Y>1]. So that means the
probabilities of 1 to 15, which is NOT zero to a number. But we know that
from 0 to 15 equals a probability of ONE, because all the probabilities must
add up to one. So is we get rid of the part of that that we do not want (which
in this case is 0) we would have the answer. So if we look up k equal to 0
(which will give us the probability of 0) we get: means I choose a k of 0,
giving an answer of: 0.206; so the answer to this problem is
1 – 0.206 = 0.794
Problem #142
“…no more than three…” is the same as saying P[Y<3]. So that means the
probabilities of 0 to 3, which means I choose a k of 3, giving an answer of:
0.944
Problem #143
(mean)  = n*p = 15*0.1 = 1.5
(standard deviation)  = √15 ∗ 0.1 ∗ 0.9 = 1.1619
Problem #144
This problem harkens back to the end of chapter 7 ( reliability of systems) in
which we have a parallel system. As you may recall the equation of a
parallel system’s reliability is: R = 1 – (1-a)(1-b)(1-c)…or if all the
reliabilities are the same then it is R = 1 – (1-a)n Since n is the number of (in
this case) radar stations, that have a reliability of detecting “…an incoming
enemy missile is only 0.4…” And we want a reliability of the system to be
“…at least 0.99…” then the equation becomes:
0.99 = 1 – (1-0.4)n (subtract 1 from both sides)
-0.01 = – (0.6)n (multiply both sides by -1)
0.01 = (0.6)n (convert to a log)
ln(0.01)/ln(0.6) = n = 9.0152
Since n = 9.0152 is for exactly 0.99 and we can only have whole numbers
of radar stations, we must go to 10 so that we have “at least 0.99”: 10
BUT there is a way of looking at this problem from a binomial
standpoint… How many radar sites are needed such that AT LEAST ONE
of them detects the enemy aircraft. Remember that “at least one” is the same
as 1 – P(0);so if I want at least 0.99 for the probability then the P(0) must
equal 0.01 or less. Using the equation for binomial probability:
P(0) = C(n,0)*0.40*0.6(n-0) = 0.6n = 0.01 (remember any combination with a
“0” equals 1 and number raised to the “0 power” equals 1)
please note that that is the same as we got above…so,
0.01 = (0.6)n (convert to a log)
ln(0.01)/ln(0.6) = n = 9.0152
Since n = 9.0152 is for exactly 0.99 and we can only have whole numbers
of radar stations, we must go to 10 so that we have “at least 0.99”: 10
Problems #145-148
Since it says “…If ten of these lights are selected selected…” then n=10; and
since it says “…will last for at least 200 hours is 0.80…” then p=0.80; and
this is what will be used with the appropriate part of table 1 to solve the
problems…
Problem #145
“…at least eight…” is the same as saying P[Y>8]. So that means the
probabilities of 8 to 10, which is NOT zero to a number. But we know that
from 0 to 10 equals a probability of ONE, because all the probabilities must
add up to one. So is we get rid of the part of that that we do not want (which
in this case is 0 to 7) we would have the answer. So if we look up k equal to
7 (which will give us the probability of 0 to 7) we get: means I choose a k of
7, giving an answer of: 0.322; so the answer to this problem is
1 – 0.322 = 0.678
Problem #146
“…at most nine…” is the same as saying P[Y<9]. So that means the
probabilities of 0 to 9, which means I choose a k of 9, giving an answer of:
0.893
Problem #147
“…between six and eight inclusive…” is the same as saying P[6<Y<8]. So
that means the probabilities of 6 to 8, which is NOT zero to a number. We
know that if we choose k equal to 8 that it is the same as:
P(0)+P(1)+P(2)+P(3)+P(4)+P(5)+P(6)+P(7)+P(8),
which has too much, it also has
P(0)+P(1)+P(2)+P(3)+P(4)+P(5), which is the same as if you chose a k equal
to 5…so if you subtract these you get the answer to this problem is
0.624 – 0.033 = 0.591
Problem #148
(mean)  = n*p = 10*0.8 = 8
(standard deviation)  = √10 ∗ 0.8 ∗ 0.2 = 1.2649
Problems #149-152
Since it says “…If we test five…” then n=5; and since it says “…defective
in 30%…” then p=0.30; and this is what will be used with the appropriate
part of table 1 to solve the problems…
Problem #149
“…at least two…” is the same as saying P[Y>2]. So that means the
probabilities of 2 to 5, which is NOT zero to a number. But we know that
from 0 to 5 equals a probability of ONE, because all the probabilities must
add up to one. So is we get rid of the part of that that we do not want (which
in this case is 0 to 1) we would have the answer. So if we look up k equal to
1 (which will give us the probability of 0 to 1) we get: means I choose a k of
1, giving an answer of: 0.528; so the answer to this problem is
1 – 0.528 = 0.472
Problem #150
“…at most two…” is the same as saying P[Y<2]. So that means the
probabilities of 0 to 2, which means I choose a k of 2, giving an answer of:
0.837
Problem #151
“expected value” is the same as the mean:  = n*p = 5*0.3 = 1.5
Problem #152
(standard deviation)  = √5 ∗ 0.3 ∗ 0.7 = 1.0247
Problems #153-154
In Section 8.3, problem 75 says “…Suppose five patients…” then n=5; and
since it says “…radiation treatment is 70% effective…” then p=0.70; and
this is what will be used with the appropriate part of table 1 to solve the
problems…
Problems #153
For the probability distribution of n=5 we must calculate each of the
probabilities from 0 to 5:
P(0) = C(5,0)*(0.7)0*(0.3)5 = 0.00243
P(1) = C(5,1)*(0.7)1*(0.3)4 = 0.02835
P(2) = C(5,2)*(0.7)2*(0.3)3 = 0.1323
P(3) = C(5,3)*(0.7)3*(0.3)2 = 0.3087
P(4) = C(5,4)*(0.7)4*(0.3)1 = 0.36015
P(5) = C(5,5)*(0.7)5*(0.3)0 = 0.16807
OR subtract the previous value in the table to get each value:
P(0) = 0.002
P(1) = 0.031 – 0.002 = 0.029
P(2) = 0.163 – 0.031 = 0.132
P(3) = 0.472 – 0.163 = 0.309
P(4) = 0.832 – 0.472 = 0.360
P(5) = 1 – 0.832 = 0.168
This confirms the match with the probability distribution of problem
#75 (with only one exception because of rounding if you calculate each
value using the equation…P(2))
Problem #154
We know that if it is a binomial the answers will be:
(mean)  = n*p = 5*0.7 = 3.5
(standard deviation)  = √5 ∗ 0.7 ∗ 0.3 = 1.0247
Now using the method we used in section 8.3:
 = 0*0.00243 + 1*0.02835 + 2*0.1323 + 3*0.3087 + 4*0.36015 + 5*0.16807
 = 3.5
E(X2) = 02*0.00243 + 12*0.02835 + 22*0.1323 + 32*0.3087 + 42*0.36015 +
52*0.16807 = 13.3
(variance) 2 = E(X2) - 2 = 13.3 – 3.52 = 1.05
(standard deviation)  = square root of 2 = 1.0247
Isn’t nice to see that they match?