Problem 210-218
For this group of exercises you will apply the equations for Combinations and Permutations and simplify, then multiply…rather than just entering into your calculator and getting an answer. The equations are: (for Permutations) 𝑃(𝑛, 𝑘) = 𝑛!
(𝑛−𝑘)!
and (for Combinations) 𝐶(𝑛, 𝑘) = 𝑛!
(𝑛−𝑘)!𝑘!
Problem 210
Combination: 𝐶(𝑛, 𝑘) = 𝑛!
(𝑛−𝑘)!𝑘!
, so 𝐶(7,3) =
7!
(7−3)!3!
=
7!
4!3!
=
7∗6∗5∗4∗3∗2∗1
4∗3∗2∗1∗ 3∗2∗1
=
7∗6∗5
3∗2∗1
= 7 ∗ 5 = 35
Problem 211
Permutation: 𝑃(𝑛, 𝑘) = 𝑛!
(𝑛−𝑘)!
, so 𝑃(14,4) =
14*13*12*11 = 24,024
14!
(14−4)!
=
14!
10!
=
14∗13∗12∗11∗10∗9∗8∗7∗6∗5∗4∗3∗2∗1
10∗9∗8∗7∗6∗5∗4∗3∗2∗1
=
Problem 212
Permutation: 𝑃(𝑛, 𝑘) = 𝑛!
(𝑛−𝑘)!
, so 𝑃(9,9) =
9!
(9−9)!
=
9!
0!
=
9∗8∗7∗6∗5∗4∗3∗2∗1
1
= 362,880
Problem 213
Combination: 𝐶(𝑛, 𝑘) = 𝑛!
(𝑛−𝑘)!𝑘!
, so 𝐶(14,4) =
14∗13∗12∗11
4∗3∗2∗1
= 7 ∗ 13 ∗ 11 = 1001
14!
(14−4)!4!
=
14!
10!4!
=
14∗13∗12∗11∗10∗9∗8∗7∗6∗5∗4∗3∗2∗1
=
10∗9∗8∗7∗6∗5∗4∗3∗2∗1∗ 4∗3∗2∗1
Problem 214
Combination: 𝐶(𝑛, 𝑘) = 𝑛!
(𝑛−𝑘)!𝑘!
, so 𝐶(18,17) =
18∗17∗16∗15∗14∗13∗12∗11∗10∗9∗8∗7∗6∗5∗4∗3∗2∗1
= 18
1∗1716∗15∗14∗13∗12∗11∗10∗9∗8∗7∗6∗5∗4∗3∗2∗1
=
18
1
18!
(18−17)!17!
=
18!
1!17!
=
Problem 215
Permutation: 𝑃(𝑛, 𝑘) = 𝑛!
(𝑛−𝑘)!
, so 𝑃(7,3) =
7!
(7−3)!
=
7!
4!
=
7∗6∗5∗4∗3∗2∗1
4∗3∗2∗1
= 7*6*5 = 210
Problem 216
Several Combinations: 𝐶(𝑛, 𝑘) = 𝑛!
(𝑛−𝑘)!𝑘!
, so
𝐶(4,3)∗𝐶(6,2)
𝐶(10,5)
=
4!
(4−3)!3!
∗
6!
(6−2)!2!
10!
=
4!
1!3!
∗
6!
4!2!
10!
(10−5)!5!
5!5!
=
4∗3∗2∗1
∗
6∗5∗4∗3∗2∗1
1∗3∗2∗1 4∗3∗2∗1∗ 2∗1
10∗9∗8∗7∗6∗5∗4∗3∗2∗1
5∗4∗3∗2∗1∗ 5∗4∗3∗2∗1
=
4
∗
6∗5
1 2∗1
10∗9∗8∗7∗6
5∗4∗3∗2∗1
=
4∗3∗5
3∗2∗7∗6
5
=
7∗3
= 5/21 or 0.2381

Problem 217
Several Combinations:
7∗6∗5∗4∗3∗2∗1
∗
4∗3∗2∗1
5∗4∗3∗2∗1∗2∗1 2∗1∗ 2∗1
11∗10∗9∗8∗7∗6∗5∗4∗3∗2∗1
7∗6∗5∗4∗3∗2∗1∗ 4∗3∗2∗1
=
𝐶(𝑛, 𝑘) = 𝑛!
(𝑛−𝑘)!𝑘!
, so
𝐶(7,2)∗𝐶(4,2)
𝐶(11,4)
=
7!
(7−2)!2!
∗
4!
(4−2)!2!
11!
=
7!
∗
4!
5!2!
2!2!
11!
(11−4)!4!
7!4!
4
∗
6∗5
1 2∗1
11∗10∗9∗8
4∗3∗2∗1
4∗3∗5
=
11∗10∗3
4
=
11∗2
= 4/22 =2/11 or 0.1818
=
Problem 218
Permutation: 𝑃(𝑛, 𝑘) = 𝑛!
(𝑛−𝑘)!
, so 𝑃(10,1) =
10!
(10−1)!
=
10!
9!
=
10∗9∗8∗7∗6∗5∗4∗3∗2∗1
9∗8∗7∗6∗5∗4∗3∗2∗1
= 10
Problem 219-229
To best prepare for these let’s put the data into a table:
Event
Probability
Set A
Set B
Set C
Problem 219 a
0.15
B b
0.20
C c
0.25
A d
0.30
A
B e
0.10
A
B
C
To find a probability of a set, add up the individual probabilities of the elements…
P[A] = 0.25 + 0.30 + 0.10 = 0.65
P[B] = 0.15 + 0.30 + 0.10 = 0.55
P[C] = 0.20 + 0.10 = 0.30
Problem 220
To find a probability of an intersection, determine the elements in the intersection (remember it is where they “overlap”) and add up the individual probabilities of the elements…
P[A∩B] = P[d] + P[e] = 0.30 + 0.10 = 0.40
P[A∩C] = P[e] = 0.10
P[B∩C] = P[e] = 0.10
Problem 221
To find a probability of a union, determine the elements in the union (remember it is anywhere you find either of the letters) and add up the individual probabilities of the elements…
P[AUB] = P[a] + P[c] + P[d] + P[e] = 0.15 + 0.25 + 0.30 + 0.10 = 0.80 [or 1-P[b] = 1 – 0.20]
P[AUC] = P[b] + P[c] + P[d] + P[e] = 0.20 + 0.25 + 0.30 + 0.10 = 0.85 [or 1-P[a] = 1 – 0.15]
P[BUC] = P[a] + P[b] + P[d] + P[e] = 0.15 + 0.20 + 0.30 + 0.10 = 0.75 [or 1-P[b] = 1 – 0.25]

Problem 222
To find a probability of a conditional probability you use the formula (P[A|B] = P[A∩B]/P[B]) which will always work:
Using the answers from 219-220, P[A|B] = P[A∩B]/P[B] = 0.40/0.55 = 8/11 or 0.7273
Using the answers from 219-220, P[A|C] = P[A∩C]/P[C] = 0.10/0.30 = 1/3 or 0.3333
Problem 223
To find a probability of a conditional probability you use the formula (P[A|B] = P[A∩B]/P[B]) which will always work:
Using the answers from 219-220, P[B|C] = P[B∩C]/P[C] = 0.10/0.30 = 1/3 or 0.3333
Using the answers from 219-220, P[B|A] = P[B∩A]/P[A] = 0.40/0.65 = 8/13 or 0.6154
Problem 224
There are lots of explanations for independence accurate way of determining that is to use the formula
(P[A|B] = P[A]); if equal then they are independent, but if not then they are NOT independent (or they are dependent):
Using the answers from 222 and 219, P[A|B] = P[A] => 0.7273 ≠ 0.65 so A and B are NOT
INDEPENDENT
Problem 225
There are lots of explanations for independence accurate way of determining that is to use the formula
(P[A|B] = P[A]); if equal then they are independent, but if not then they are NOT independent (or they are dependent):
Using the answers from 222 and 219, P[A|C] = P[A] => 0.3333 ≠ 0.65, so A and C are NOT
INDEPENDENT
Problem 226
There are lots of explanations for independence accurate way of determining that is to use the formula
(P[A|B] = P[A]); if equal then they are independent, but if not then they are NOT independent (or they are dependent):
Using the answer from 223 and 219, P[B|C] = P[B] => 0.3333 ≠ 0.55, so B and C are NOT
INDEPENDENT
Problem 227
Mutually exclusive simply comes down to the question of whether there is an intersection…if there is
NO INTERSECTION then they are Mutually Exclusive.
Using the answers from 220, P[A∩B] = 0.40 ≠ 0, so A and B are NOT Mutually Exclusive

Problem 228
Mutually exclusive simply comes down to the question of whether there is an intersection…if there is
NO INTERSECTION then they are Mutually Exclusive.
Using the answers from 220, P[A∩C] = 0.10 ≠ 0, so A and C are NOT Mutually Exclusive
Problem 229
Mutually exclusive simply comes down to the question of whether there is an intersection…if there is
NO INTERSECTION then they are Mutually Exclusive.
Using the answers from 220, P[B∩C] = 0.10 ≠ 0, so B and C are NOT Mutually Exclusive
Problem 230
Looking for the intersection…and that can be found in the Union (or Additive Law) equation or in the
Conditional Probability equation. Since there are fewer “pieces” in the Conditional Probability equation we will look at that first: P[A|B] = P[A∩B]/P[B] = 0.4 = P[A∩B]/0.3
Therefore, P[A∩B] = 0.12
Problem 231
Looking for the P[A] leads us to the Union (or Additive Law) equation: P[AUB] = P[A] + P[B] - P[A∩B] =
0.5 = P[A] + 0.3 – 0.12
Therefore, P[A] = 0.32
Problem 232
P[B|A] = P[B∩A]/P[A] = 0.12/0.32
Therefore, P[B|A] = 3/8 or 0.375
Problem 233
There are lots of explanations for independence accurate way of determining that is to use the formula
(P[A|B] = P[A]); if equal then they are independent, but if not then they are NOT independent (or they are dependent):
Using the the given for P[A|B] and the answer from 231, P[A|B] = P[A] => 0.4 ≠ 0.32, so A and B are
NOT INDEPENDENT
Problem 234
Mutually exclusive simply comes down to the question of whether there is an intersection…if there is
NO INTERSECTION then they are Mutually Exclusive.
Using the answers from 230, P[A∩B] = 0.12 ≠ 0, so A and B are NOT Mutually Exclusive

Problem 235
It says “…C and D are independent…”, which means that P[C] = P[C|D] = 0.4 (given). So that means that P[C∩D]/P[D] = 0.4, or P[C∩D] = 0.4*P[D]. And why would I do that? Because I know that the only other piece of given information is a union which has an equation with both P[D] and P[C∩D], and we are looking to solve for P[D]…so I plan to substitute into P[CUD] = P[C] + P[D] – P[C∩D], so it becomes:
0.6 = 0.4 + P[D] -0.4*P[D] => 0.2 = 0.6*P[D]
So, P[D] = 1/3 or 0.3333
Problem 236
Starting key phrase “…are selected…”, which is another way of that someone chose them… so we will use combinations (order is not required). So for this case you must choose 2 Harvards (out of 7 total
Harvards) and 1 Yale (out of a total of 4 Yales); (the 1 Yale is since you need a total of 3):
C(7,2)*C(4,1) = 21 *4 = 84 possible ways…but not probability; that requires that I divided by the total number of outcomes (choosing 3 people out of a total of 11): C(11,3) = 165
So the probability is 84/165 = 28/53 or 0.xxxx
Problem 237
Starting key phrase “…are selected…”, which is another way of that someone chose them… so we will use combinations (order is not required). Next phase of importance is “…at least one…”, which means
1 or more, so in this case we want 1 Yale, 2 Yales or 3 Yales…or in other words all except for 0 Yales (or all Harvard). So you can either calculate 1, 2 and 3 Yales and add them up OR figure the probability of all Harvards and subtract from 1 (complement rule)…I will do it both ways.
First, 1 Yale is the same as problem 236, but 2 Yales will be to choose 2 Yales (out of 4 total Yales) and 1
Harvard (out of a total of 7 Harvards); (the 1 Harvard is since you need a total of 3):
C(4,2)*C(7,1) = 6 *7 = 42 possible ways; also for 3 Yales it is C(4,3) = 4…but not probability; that requires that I divided by the total number of outcomes (choosing 3 people out of a total of 11):
C(11,3) = 165
So the probability is (84+42+4)/165 = 130/165 = 26/35 or 0.xxxx
Similarly, 3 Harvards it is C(7,3) = 35…but not probability; that requires that I divided by the total number of outcomes (choosing 3 people out of a total of 11): C(11,3) = 165
So the probability is 1-35/165 = 130/165 = 26/33 or 0.xxxx
Problem 238
A subpart to problem 237… 3 Harvards it is C(7,3) = 35…but not probability; that requires that I divided by the total number of outcomes (choosing 3 people out of a total of 11): C(11,3) = 165

So the probability is 35/165 = 7/33 or 0.xxxx
Problem 239
Starting key phrase “…are selected…”, which is another way of that someone chose them… so we will use combinations (order is not required). Next there are two sets of choices to be made (because there are two packages to choose from)… One last thing to consider is there are kinds of batteries
(good and defective) in each package. Finally all this must be done together making one complete choice.
So, next it says “…at least three of the four…”, so either 3 OR 4 good batteries. For 4, that is easy, all the chosen must be good. But for 3, either the first has a defective and not the second, or the other way around.
All 4: Choosing 2 from the first pack (having 4 good batteries) and choosing 2 from the second pack
(having 3 good batteries): C(4,2)*C(3,2) = 6*3 = 18
3 (1 st pack has defective): Choosing 1 from the first pack (having 4 good batteries), choosing 1 defective from the first pack (having 2 defective) and choosing 2 from the second pack (having 3 good batteries):
C(4,1)*C(2,1)*C(3,2) = 4*2*3 = 24
3 (2 nd pack has defective): Choosing 2 from the first pack (having 4 good batteries), choosing 1 defective from the second pack (having 3 defective) and choosing 1 from the second pack (having 3 good batteries): C(4,2)*C(3,1)*C(3,2) = 6*3*3 = 54
Lastly all possible choices for both packages (because we have to have the total outcomes to create a probability: C(6,2)*C(6,2) = 15*15 = 225
So the probability is (18+24+54)/225 = 96/225 = 32/75 or 0.xxxx
Problem 240
A subpart to problem 239… 4 good batteries, it is C(4,2)*C(3,2) = 6*3 = 18…but not probability; that requires that I divided by the total number of outcomes (choosing 2 from each pack of 6:
C(6,2)*C(6,2) = 225
So the probability is 18/225 = 2/25 or 0.08
Problem 241
A variation to problem 239… choosing one good from the 1 st pack (out of 4) and one bad battery (out of 2), as well as choosing, one good from the 2 nd pack (out of 3) and one bad battery (out of 3):
C(4,1)*C(2,1)*C(3,1)*C(3,1) = 4*2*3*3 = 72…but not probability; that requires that I divided by the total number of outcomes (choosing 2 from each pack of 6: C(6,2)*C(6,2) = 225
So the probability is 72/225 = 8/25 or 0.16
Problem 242

“A series system…” should bring to mind to system reliability equations: R = a*b*c (for a series system) and R = 1 – (1-a)*(1-b)*(1-c) (for a parallel system).
So, you have “…three B’s in a parallel arrangement, and one A.” Should look like this:
add in the reliabilities>
7
0.7 0.9
0.7
This is obviously a combination of a series and a parallel (it even says it in the text). So we need to do one part at a time and simplify, so start with the obvious parallel part and use the equation:
R = 1 – (1 - 0.7)*(1 – 0.7)*(1 – 0.7) = 1 – 0.3
3 = 0.973
Which converts it to: 0.97
0.9
3
And since this is a simple series the answer is: R = 0.973*0.9 = 0.8757
Problem 243
“A series system…” should bring to mind to system reliability equations: R = a*b*c (for a series system) and R = 1 – (1-a)*(1-b)*(1-c) (for a parallel system).
So, you have “…four B components in parallel and three A components in parallel.” Should look like this:
0.7
add in the reliabilities>
0.7
0.7
0.7
0.9
0.9
0.9
This is obviously a combination of a series and 2 parallels (it even says it in the text). So we need to do one part at a time and simplify, so start with the first parallel part and use the equation:
R = 1 – (1 - 0.7)*(1 – 0.7)*(1 – 0.7)*(1– 0.7) = 1 – 0.3
4 = 0.9919
Then the second parallel: R = 1 – (1 - 0.9)*(1 – 0.9)*(1 – 0.9) = 1 – 0.9
3 = 0.999
Which converts it to: 0.991
0.999
9
And since this is a simple series the answer is: R = 0.9919*0.999 = 0.9909
Problem 244
FCP…2 tasks: rolling the die (assumed 6-sided) and flipping Delmo.

Since it says “…Delmo gets a 3 or Delmo lands on his tail.” then it is a union of all events that have either Delmo rolling a 3 or him landing on tails, both. So, 3 and tails, not a 3 but tails, and 3 but not a tail would have to be calculated and added together. Or do the complement, figure for what does not match and subtract from 1…no 3 and no tails. It will be done both ways:
3 + tails: 1/6 * 1/2 = 1/12 no 3 + tails: 5/6 * 1/2 = 5/12
3 + no tails: 1/6 * 1/2 = 1/12
So the answer is 1/12 + 5/12 + 1/12 = 7/12 or 0.5833 no 3 + no tail: 5/6 * 1/2 = 5/12
So the answer is 1 -5/12 = 7/12 or 0.5833
Problem 245
This is a subset of problem 244…just the part with getting both a 3 and a tail:
3 + tails: 1/6 * 1/2 = 1/12 or 0.0833
Problem 246
“Component A has reliability…” should bring to mind to system reliability equations: R = a*b*c (for a series system) and R = 1 – (1-a)*(1-b)*(1-c) (for a parallel system).
So, you have “…four such components [R = 0.6] along with two transistors [R = 0.9]…components are in parallel.” Should look like this:
0.6
add in the reliabilities>
0.6
0.9
0.6 0.9
This is obviously a combination of a series and 2 parallels (it even says it in the text). So we need to do one part at a time and simplify, so start with the first parallel part and use the equation:
R = 1 – (1 - 0.6)*(1 – 0.6)*(1 – 0.6)*(1 – 0.6) = 1 – 0.4
4 = 0.9744
Then the second parallel: R = 1 – (1 - 0.9)*(1 – 0.9) = 1 – 0.9
2 = 0.99
Which converts it to: 0.9744 0.99
And since this is a simple series the answer is: R = 0.9744*0.99 = 0.9647
Problem 247

By now cards should signify combinations… choosing 3 twos (out of the 4 available), then choosing the jack of hearts (out of the 1 jack of hearts available) and lastly choosing 1 card out of all the other
“non-twos/jacks available): C(4,3)*C(1,1)*C(44,1) = 4*1*44 = 176…but not probability; that requires that I divided by the total number of outcomes (choosing 5 from 52 cards: C(52,5) = 2,598,960
So the probability is 176/2598960 = 0.00007
Problem 248
By now cards should signify combinations… choosing 1 three (out of the 4 available), then choosing 2 fours (out of the 4 available) and lastly choosing 2 red sevens (out of only 2 available, remember you only want the red ones: hearts and diamonds, not the spades or clubs): C(4,1)*C(4,2)*C(4,2) = 4*6*6 =
144…but not probability; that requires that I divided by the total number of outcomes (choosing 5 from
52 cards: C(52,5) = 2,598,960
So the probability is 144/2598960 = 0.00005
Problem 249
In the problem, near the question, it states “...probability of successfully opening the safe, if he gets past the alarm…” shows us a conditional probability. So the equation P[A|B] = P[A∩B]/P[B] applies.
We are given that “..the burglar gets past the alarm and opens the safe is 0.6 .” Which (remember
“and” is an intersection) means P[A∩B] = 0.6 . We are also told that the “...probability of successfully opening the safe, if he gets past the alarm…” is 0.8. Which (remember “if” the conditional probability
“|”) means P[B|A] = 0.8 . So P[B|A] = P[B∩A]/P[A] becomes 0.8 = 0.6/P[A]
So the probability of getting past the alarm is P[A] = 0.6/0.8 = 6/8 = ¾ or 0.75
Problem 250
In this problem we are arranging flags in an order and we have multiples of the same color, so we are going to do things like we did in Section 7.2.
If all the flags were a different color, then there would be 8! Different ways to arrange the eight flags.
BUT the duplication of colors means we cannot tell a difference if we change (say) to blue flags. You might also remember the problem of words like TENNESSEE…well this is the “word” BBBBGGGW”.
So you get 8!/(4!*3!*1!) = 40320/(24*6*1) = 40320/144 = 280
Problem 251
This is a combinations problem…not because the sergeant is choosing, but because the recruits are choosing (whether to volunteer or not)…and if it was the military really, you would expect 6 every time because they are “brown-nosing”).
In any case, it says “…at least 4…” which means 4 or more, so 4, 5 or 6 volunteers. Hmmm, “or” means union so we will be adding these answers together.

4 volunteers out 6 recruits: C(6,4) = 21
5 volunteers out 6 recruits: C(6,5) = 6
6 volunteers out 6 recruits: C(6,6) = 1
So the answer is 21 + 6 + 1 = 28
Problem 252-258
Given a table it is time to fill it out with the totals of the columns and rows to make our work easier when addressing the problems.
Studied
Did Not
Study Total
Passed Exam
Did Not Pass
290
40
60
110
350
150
Total 330 170 500
Also remember that probability is how many answer the question divided by the total.
Problem 252
“…studied for the exam…” has 330 out of a total of 500…so the answer is 330/500 = 33/50 or 0.66
Problem 253
“…passed the exam…” has 350 out of a total of 500…so the answer is 350/500 = 7/10 or 0.7
Problem 254
“…we know the student selected passed the first exam.” Which means that the total is no longer the total of 500, but only the 350 that passed (this is a conditional probability). Now the question has to do with whether the student studied…and only 290 of the 350 studied… so the answer is 290/350 = 29/35 or 0.8286
Problem 255
“…we know the student did not study...” which means that the total is no longer the total of 500, but only the 170 that did not study (this is a conditional probability). Now the question has to do with whether the student passed…and only 60 of the 170 passed… so the answer is 60/170 = 6/17 or 0.3529
Problem 256
“…we know the student studied...” which means that the total is no longer the total of 500, but only the 330 that did study (this is a conditional probability). Now the question has to do with whether the student passed…and only 290 of the 330 passed… so the answer is 290/330 = 29/33 or 0.8788
Problem 257

“…mutually exclusive...” which means that they do not overlap in any way (or there is no intersection).
But they do intersect (there are students that have both studied and passed the exam), so they are
NOT Mutually Exclusive
Problem 258
“…independent...” which means is the probability affected if you are given some information (or does the P[studied] = P[studied|passed]). P[studied] is answered in problem 252, P[studied|passed] is answered in problem 254. So then we know that 0.66 ≠ 0.8286, so they are NOT Independent
Problem 259-261
Looking at the questions I notice the word“…if...” in the middle of the questions…this says that conditional probability will be used. So then the equation P[A|B] = P[A∩B]/P[B] will be used and a tree diagram or table will be very helpful…so let’s generate both first.
There are two activities in this…having a milkshake (or not) and sleeping well (or not)…and for the tree diagram it is a matter of timing, so the milkshake happened before the sleeping (in this case)…
Table: Tree:
G
M
Good Rest
Poor Rest
Total
Milkshake
No
Milkshake Total
N
P
G
P
To finish and fill these in…we know that “When Delmo has a milkshake…only a 50% chance of a good night’s sleep.” (Which means a 50% chance of a poor night’s sleep.)
Table: Tree:
0.50 G
M
Milkshake
Good Rest 0.50*M
Poor Rest
Total
No
Milkshake Total
N
0.50 P
G
P
Next…we know that “…doesn’t have a milkshake…only a 80% chance of a good night’s sleep.” (Which means a 20% chance of a poor night’s sleep.)
Table: Tree:
0.50 G
M
Milkshake
No
Milkshake Total
0.50 P
N
0.80
0.20
G
P

Good Rest 0.50*M 0.80*N
Poor Rest 0.50*M 0.20*N
Total
Lastly, we know that “…has a milkshake [M] 60% of the time...” (Which means a 40% of the time he doesn’t. [N]) Just the totaling and multiplying the paths to finish these two…
Table: Tree:
0.50 G 0.30
M
Good Rest
Poor Rest
Total
Milkshake
0.30
0.30
0.60
No
Milkshake
0.32
0.08
0.40
Total
0.62
0.38
1.00
0.60
0.40
N
0.50
0.80
0.20
P
G
P
0.30
0.32
0.08
Now to solve the questions…
Problem 259
“…good night’s sleep.” which means all those paths that have a good night’s sleep (tree), or from the table, the total for good night’s sleep… So the answer is 0.62
Problem 260
“…milkshake before bedtime if we know that he slept poorly.” The “if” says it is a conditional probability that looks like P[M|P], which equals P[M∩P]/P[P]. P[M∩P] is the intersection of Milkshake and Poor sleep…on the table it is 0.30, which on the tree is the path of M and P (still 0.30). P[P] is the probability of poor sleep (which is the opposite or complement to the answer in problem 259, so it is 1
– 0.62 = 0.38.
So the answer is 0.30/0.38 = 30/38 = 15/19 or 0.7895
Problem 261
“…did not have a milkshake before bedtime if he slept well...” The “if” says it is a conditional probability that looks like P[N|G], which equals P[N∩G]/P[G]. P[N∩G] is the intersection of No milkshake and Good sleep…on the table it is 0.32, which on the tree is the path of N and G (still 0.32).
P[G] is the probability of good sleep which is the answer in problem 259.
So the answer is 0.32/0.62 = 32/62 = 16/31 or 0.5161
Problem 262
“Find the reliability…” should bring to mind to system reliability equations: R = a*b*c (for a series system) and R = 1 – (1-a)*(1-b)*(1-c) (for a parallel system).

So, you have the following:
0.84
0.65 0.92
0.84
This is obviously a combination of a series and a parallel. So we need to do one part at a time and simplify, so start with the parallel part first and use the equation:
R = 1 – (1 - 0.84)*(1 – 0.84) = 1 – 0.16
2 = 0.9744
Which converts it to:
0.9744 0.65
0.92
Problem 263-269
Looking at the questions I notice the word“…if...” in the middle of the questions…this says that conditional probability will be used. So then the equation P[A|B] = P[A∩B]/P[B] will be used and a tree diagram or table will be very helpful…so let’s generate both first.
There are two activities in this…Small Budget (or Large Budget) and Small Sales (or Large Sales or
Average Sales)…and for the tree diagram it is a matter of timing, so the budget happens before the sales…
SS
Table: Tree: SB
LS
AS
Small $$
Large $$
Average $$
Total
Small
Budget
Large
Budget Total
LB
SS
LS
AS
Next, we know that for a “…small budget, there is a 40% chance of having large sales, a 30% chance of having average sales, and a 30% chance of poor sales…”
Table:
Small $$
Large $$
Small
Budget
0.30*SB
0.40*SB
Large
Budget Total
Tree:
SB
0.30
0.30
0.40
SS
LS
SS
AS
LB AS
LS

Average $$ 0.30*SB
Total
Next, we know that for a “…large advertising budget, there is a 65% chance of having large sales, a
25% chance of having average sales, and only 10% chance of poor sales.”
Table: Tree:
SB
0.30
0.30
SS
AS
Small
Budget
Large
Budget
Small $$
Large $$
0.30*SB
0.40*SB
0.10*LB
0.65*LB
Average $$ 0.30*SB 0.25*LB
Total
Total
LB
0.40
0.25
0.10
0.65
LS
SS
LS
AS
Lastly, we know that “…use a large advertising budget [LB] 40% of the time...” (Which means a 60% of the time it’s a small budget. [SB]) Just the totaling and multiplying the paths to finish these two…
Table:
Small
Budget
0.18 Small $$
Large $$ 0.24
Average $$ 0.18
Total 0.60
Large
Budget
0.04
0.26
0.10
0.40
Total
0.22
0.50
0.28
1.00
Tree:
0.60
0.40
SB
LB
0.30
0.30
0.40
0.25
0.10
0.65
SS
LS
SS
LS
0.18
AS
0.24
0.04
AS
0.26
0.18
0.10
Now to solve the questions…
Problem 263
“…has a large sales ...” which means all those paths that have a large sales (tree), or from the table, the total for Large $$… So the answer is 0.50
Problem 264
“…has an average sales ...” which means all those paths that have an average sales (tree), or from the table, the total for Average $$… So the answer is 0.28
Problem 265
“…has a poor sales ...” which means all those paths that have a poor sales (tree), or from the table, the total for Poor $$… So the answer is 0.22
Problem 266

“…large advertising budget if we learn they had a large sales month.” The “if” says it is a conditional probability that looks like P[LB|LS], which equals P[LB∩LS]/P[LS]. P[LB∩LS] is the intersection of Large
Budget and Large Sales…on the table it is 0.26, which on the tree is the path of LB and LS (still 0.26).
P[LS] is the probability of large sales…from problem 263.
So the answer is 0.26/0.50 = 26/50 = 13/25 or 0.52
Problem 267
“…small advertising budget if we learn they had a poor sales month.” The “if” says it is a conditional probability that looks like P[SB|PS], which equals P[SB∩SS]/P[SS]. P[SB∩SS] is the intersection of Small
Budget and Small Sales…on the table it is 0.18, which on the tree is the path of SB and SS (still 0.18).
P[SS] is the probability of large sales…from problem 265.
So the answer is 0.18/0.22 = 18/22 = 9/11 or 0.8182
Problem 268
“…small advertising budget if we learn they had a large sales month.” The “if” says it is a conditional probability that looks like P[SB|LS], which equals P[SB∩LS]/P[LS]. P[SB∩LS] is the intersection of Small
Budget and Small Sales…on the table it is 0.24, which on the tree is the path of SB and LS (still 0.18).
P[LS] is the probability of large sales…from problem 263.
So the answer is 0.24/0.50 = 24/50 = 12/25 or 0.48
Problem 269
“…small advertising budget if we learn they had a average sales month.” The “if” says it is a conditional probability that looks like P[SB|AS], which equals P[SB∩AS]/P[AS]. P[SB∩AS] is the intersection of Small
Budget and Small Sales…on the table it is 0.18, which on the tree is the path of SB and AS (still 0.18).
P[AS] is the probability of large sales…from problem 264.
So the answer is 0.18/0.28 = 18/28 = 9/14 or 0.6429
Problem 270
“…arrangements...” think FCP. Each task is a place in the line. With “…no restriction…” then any of the
12 people (6 couples…6*2 = 12) can be first, only 11 possible choices for the second position and so forth giving us:
12*11*10*9*8*7*6*5*4*3*2*1 = 12! = 479,001,600
Problem 271
“…arrangements...” think FCP. Each task is a place in the line. With “…couples together, women proceeding men…” then any of the 6 couples can be first, only 5 possible choices for the second position and so forth giving us: 6*5*4*3*2*1 = 6! = 720
Problem 272

“…arrangements...” think FCP. Each task is a place in the line. With “…women first in a group…” then any of the 6 women can be first, only 5 possible choices for the second position and so forth until all the women are lined up, then in the seventh position you can choose any of the 6 men, and so forth, giving us:
6*5*4*3*2*1*6*5*4*3*2*1 = 6!*6! = 720*720= 518,400
Problem 273
“…arrangements...” think FCP. Each task is a place in the line. With “…couples together…” this sounds a lot like problem 271, but in this case either the man or the woman may be first…so, then any of the 12 people (6 couples) can be first, then only the person who is the partner to the first person; now the 3 rd position can now be filled with any of the remaining 10 people, leaving the 4 th position to be the partner of the 3 rd position, and so forth, giving us: 12*1*10*1*8*1*6*1*4*1*2*1 = 46,080
Problem 274
This will be a matter of combinations because the inspector must choose 5 from the 12 to be opened
(and order does not matter). With “…a sample size of 5…” and no other restrictions, it is a matter of choosing 5 out of 12, giving us: C(12,5) = 792
Problem 275
This will be a matter of combinations because the inspector must choose 5 from the 12 to be opened
(and order does not matter). With “…only good cans…”, it is a matter of choosing 5 out of 9 (because there are 3 contaminated cans out of the 12), giving us: C(9,5) = 126
Problem 275
This will be a matter of combinations because the inspector must choose 5 from the 12 to be opened
(and order does not matter). With “…at least one bad can…”; “at least one” means one or more, so in this case 1 bad can, or 2 bad cans or 3 bad cans (or if you like the complement rule, everything except all good cans)…so let’s do it both ways:
1 bad can out of 3 possible (which means the other 4 cans must be good out of 9 possible):
C(3,1)*C(9,4) = 3*126 = 378
2 bad cans out of 3 possible (which means the other 3 cans must be good out of 9 possible):
C(3,2)*C(9,3) = 3*84 = 252
3 bad cans out of 3 possible (which means the other 2 cans must be good out of 9 possible):
C(3,3)*C(9,2) = 1*36 = 36
So the answer is: 378+252+36 = 666
Or from problems 274 and 275, we subtract all the good from all possible outcomes, giving us:
792 – 126 = 666
Problem 277

3 rd game
Lose
Win
4 th game
Lose
Looking at the answers to problems 275 and 276, you can note that there are definitely more ways to
find a contaminated can than not (666 ways to find a contaminated can vs. 126 ways not to), so it is
more likely to discover that the case is contaminated. It should be noted that in the text it asks “Is the inspector more likely…”, and that implies a probability, so to turn our numbers into probabilities you divided by the total possible outcomes (problem 274), giving you 666/792 vs. 126/792 (0.8409 vs.
0.1591 or 37/44 vs. 7/44), but either way is acceptable as proof.
Problem 278
Sounds like a tree will be the easiest to visualize and work out all the possibilities or “…different paths…”, so:
Win
You now have $6
1 st game
Win
====  2 nd game
You now have $3
You now have $7
Win
==== 
You now have $2 (You are done)
Lose
Win Lose Lose
Lose
You now have $0 (you are done)
Win Win
Lose
You have $4
==== 
You have $4
Win
You have $8 (You are done …$3 up)
Lose
Win
Lose
Win
Lose
Win
Lose
You now have $1 (You are done)
Win
Lose
You have $5
You have $1 (You are done) ==== 
Win
You have $5

Win
5 th game
(last game)
Lose
Win
Lose
Win
Lose
Win
Lose
Win
Lose
Lose
Win
Lose
Win
Lose
Win
Now count the paths (I will list them first):
{Lose, Win-Lose-Lose, Win-Win-Win, Win-Lose-Win-Lose, Win-Win-Lose-Lose, Win-Lose-Win-Win-Lose,
Win-Lose-Win-Win-Win, Win-Win-Lose-Win-Lose, Win-Win-Lose-Win-Win}
Answer: 9 possible paths
Problem 279
Sounds like a tree will be the easiest to visualize and work out all the possibilities or “…different paths…”, so:
Win
You now have $5
Win
You now have $2
(You are done)
1 st game ====  2 nd game ==== 
Lose
Win
3 rd game
Lose
Win
Lose
Win
Lose
Lose
==== 
You have $3
Win
You have $7
Win
Lose
Lose
Win

4 th game
You have $0 (you are done)
You have $4 ==== 
You have $4
You have $8 (You are done)
Win Lose
Win
Lose
5 th game
(last game)
Lose
Lose
Win Win
Win
Lose Lose
Lose
Win
Win
Now count the paths (I will list them first):
{Lose, Win-Lose, Win-Win-Win-Win, Win-Win-Lose-Lose, Win-Win-Win-Lose-Lose, Win-Win-Win-Lose-
Win,
Win-Win-Lose-Win-Win, Win-Win-Lose-Win-Lose}
Answer: 8 possible paths
Problem 280
“…arranged on a shelf...” think FCP. The tasks are to determine the arrangement of the subjects and then the arrangement of the books for each subject; (note the and we will be multiplying since it is the union of these tasks). First the 3 subjects: any of three could be first, but only 2 possibilities for the second subject, leaving the last subject; (3*2*1). Next the books in each subject: 4 math books
(4*3*2*1), 3 chemistry books (3*2*1) and finally 5 biology books (5*4*3*2*1). So the answer is:
3*2*1* 4*3*2*1* 3*2*1 *5*4*3*2*1 = 103,680
Problem 281
This is a combinations problem based on the idea of which television sets are chosen. Since we are asked to “choose” 4 TV sets with “…no defects…”, they must be chosen from the 9 TV sets that are good: C(9,4)= 126

But since we asked for a probability (and not how ways this can be chosen), we must divide by the total number of ways to choose 4 TVs out of all 12: C(12,4) = 495, giving an answer of: 126/495 = 14/55 or
0.2546
Problem 282
This is a combinations problem based on the idea of which television sets are chosen. Since we are asked to “choose” 4 TV sets with “…at least one defect…”, and “at least one” means one or more, so one defective TV, 2 defective TVs or 3 defective TVs (there are only 3 defective TVs available, otherwise you would include 4 defective TVs). Of course, there is the complement rule which says all but all good ones (from problem 281). Let’s do it both ways:
1 defective TV (out of 3) and 3 good TVs (out of 9): C(3,1)*C(9,3) = 3*84 = 252
2 defective TVs (out of 3) and 2 good TVs (out of 9): C(3,2)*C(9,2) = 3*36 = 108
3 defective TVs (out of 3) and 1 good TV (out of 9): C(3,3)*C(9,1) = 1*9 = 9 so you get a total of 252+108+9 = 369
But since we asked for a probability (and not how ways this can be chosen), we must divide by the total number of ways to choose 4 TVs out of all 12: C(12,4) = 495, giving an answer of: 369/495 = 41/55 or
0.7455
Or the complement rule says: 1 – 14/55 = 41/55 or 0.7455
Problem 283
This is a combinations problem based on the idea of which television sets are chosen. Since we are asked to “choose” 4 TV sets with “…all three defective…”, they must be chosen from the 3 TV sets that are defective and the remaining one chosen from the 9 good ones: C(3,3)*C(9,1)= 1*9 = 9
But since we asked for a probability (and not how ways this can be chosen), we must divide by the total number of ways to choose 4 TVs out of all 12: C(12,4) = 495, giving an answer of: 9/495 = 1/55 or
0.0182
Problem 284
This is a combinations problem based on the idea of choosing 6 students to go. Since the question is about
“… two of the students insist on not being separated…”, then if they do not go, there is only one choice
(the other 6), but if they do go that leaves only choosing the other 4 to go from the remaining 6: 1 +
C(6,4)= 1+6 = 7
Problem 285
This is a combinations problem based on the idea of choosing 6 students to go. Since the question is about
“… two of the students …cannot be both sent…”, then if they both stay, there is only one choice (the

other 6), but if one goes, then the other will stay and so that leaves only choosing the other 5 to go from the remaining 6; and finally if the other goes, a similar choice of 5 out 6 is made: 1 + C(6,5) +
C(6,5)= 1+6+6 = 13
Problem 286
Don’t forget the table on page 35. Since this probability we will be dividing by 36 (the total number of outcomes for throwing a pair of 6-sided dice). “… both dice show even…”,
1
2
3
4
5
6
1 2 x x x
3 x x there are 9 ways that can happen so the answer is: 9/36 = ¼ or 0.25 x
4
Problem 287
5 6 x x x
Don’t forget the table on page 35. Since this probability we will be dividing by 36 (the total number of outcomes for throwing a pair of 6-sided dice). “… at least one of the two dice shows even…”,
1
2
3
4
5
6
Problem 287
1
X
X
X
2
X
X
X x
X x
3
X
X
X
4
X
X
X
X
X
X there are 27 ways that can happen so the answer is: 27/36 = 3/4 or 0.75 x
5 x
X
6
X
X
X
X
X
X
Don’t forget the table on page 35. Since this probability we will be dividing by 36 (the total number of outcomes for throwing a pair of 6-sided dice). “… exactly one die shows an even…”,
1
2
3
4
5
6
X
X
1
X
X
X
2
X
X
X
3
X
X
X
4
X x
X
5 x
X
X
6
X

there are 27 ways that can happen so the answer is: 18/36 = 1/2 or 0.50
Problem 289
This is a combinations problem based on the idea of choosing which 3 will get the tickets out of the
5(C(5,3) = 10 total possible outcomes). Since the question is about “… Alpo and Bozo get tickets…”, then there is only 1 ticket for 3 people: C(3,1) = 3 And since this is a probability, we divide by the total:
3/10 or 0.30
Problem 290
This is a combinations problem based on the idea of choosing which 3 will get the tickets out of the
5(C(5,3) = 10 total possible outcomes). Since the question is about “… Alpo and Bozo…or…Carp and
Deltoid…or Carp and Enamel…”, then first we realize that the “or” is union and we will add the three answers together; next for any pair there is only 1 ticket for 3 people: C(3,1) = 3 And since this is a probability, we divide by the total: (3+3+3)/10 = 9/10 or 0.90
Problem 291
This is a combinations problem based on the idea of choosing which claim goes to which adjustor. So let’s do this as if we were handing them out…the first adjustor will be given 6 of the 27 claims, then the next adjustor will be given 6 of the remaining 21 claims, and the next adjustor will be given 5 of the remaining 15 claims, and the next adjustor will be given 5 of the remaining 10 claims, and the last adjustor will be given 5 of the remaining 5 claims: C(27,6)*C(21,6)*C(15,5)*C(10,5)*C(5,5) =
296010*54264*3003*252*1 = 1,215,534,490,000,000