STAT213 Worksheet #3
Solutions
Question 1
The sample
space is:
1: all on
same floor
2: 6 on one
floor; 1 on another
3: 5 on one
floor; 2 on another
4: 5 on one
floor; 2 on 2 separate floors
5: 4 on one
floor; 3 on another
6: 4 on one
floor; 2 on another floor; 1 on another floor
7: 4 on one
floor; 3 on 3 separate floors
8: 3 on one
floor; 3 on another floor; 1 on another floor
9: 3 on one
floor; 2 on another floor; 2 on another floor
10: 3 on
one floor; 2 on another floor; 2 on 2 separate floors
11: 3 on
one floor; 4 on 4 separate floors
12: 2 on
one floor; 2 on another floor; 2 on another floor; 1 on another floor
13: 2 on
one floor; 2 on another floor; 3 on 3 separate floors
14: 2 on
one floor; 5 on 5 separate floors
15: all on
separate floors
Question 2
The total
number of choices is 127 = 35,831,808.
The
strategy is to choose the floors based on how many groups of people there are
and then arrange the floors among the people.
1: all on
same floor
1 group of 7 people. Total = 12C1 = 12
2: 6 on one
floor; 1 on another
1 group of
6 people; 1 group of 1 person
12C1 x 11C1
x 7!/6! = 12 x 11 x 7 = 924
3: 5 on one
floor; 2 on another
1 group of
5 people; 1 group of 2 people
12C1 x 11C1
x 7!/5!2! = 12 x 11 x 21 = 2,772
4: 5 on one
floor; 2 on 2 separate floors
1 group of
5 people; 2 groups of 1 person
12C1 x 11C2
x 7!/5! = 12 x 55 x 42 = 27,720
5: 4 on one
floor; 3 on another
1 group of
4 people; 1 group of 1 people
12C1 x 11C1
x 7!/3!4! = 12 x 11 x 35 = 4,620
6: 4 on one
floor; 2 on another floor; 1 on another floor
1 group of
4 people; 1 group of 2 people; 1 group of 1 person
12C1 x 11C1
x 10C1 x 7!/4!2! = 12 x 11 x 10 x 105 = 138,600
7: 4 on one
floor; 3 on 3 separate floors
1 group of
4 people; 3 groups of 1 person
12C1 x 11C3
x 7!/4! = 12 x 165 x 210 = 415,800
8: 3 on one
floor; 3 on another floor; 1 on another floor
2 groups of
3 people; 1 group of 1 person
12C2 x 10C1
x 7!/3!3! = 66 x 10 x 140 = 92,400
9: 3 on one
floor; 2 on another floor; 2 on another floor
1 group of
3 people; 2 groups of 2 people
12C1 x 11C2
x 7!/3!2!2! = 12 x 55 x 210 = 138,600
10: 3 on
one floor; 2 on another floor; 2 on 2 separate floors
1 group of
3 people; 1 group of 2 people; 2 groups of 1 person
12C1 x 11C1
x 10C2 x 7!/3!2! = 12 x 11 x 45 x 420 = 2,494,800
11: 3 on
one floor; 4 on 4 separate floors
1 group of
3 people; 4 groups of 1 person
12C1 x 11C4
x 7!/3! = 12 x 330 x 840 = 3,326,400
12: 2 on
one floor; 2 on another floor; 2 on another floor; 1 on another floor
3 groups of
2 people; 1 group of 1 person
12C3 x 9C1
x 7!/2!2!2! = 220 x 9 x 630 = 1,247,400
13: 2 on
one floor; 2 on another floor; 3 on 3 separate floors
2 groups of
2 people; 3 groups of 1 person
12C2 x 10C3
x 7!/2!2! = 66 x 120 x 1260 = 9,979,200
14: 2 on
one floor; 5 on 5 separate floors
1 group of
2 people; 5 groups of 1 person
12C1 x 11C5
x 7!/2! = 12 x 462 x 2520 = 13,970,880
15: all on
separate floors
12P7 =
3,991,680
Now for the
frequency table:
|
Event |
Frequency |
Percent |
|
case
1 |
12 |
0.00003% |
|
case
2 |
924 |
0.00258% |
|
case
3 |
2,772 |
0.00774% |
|
case
4 |
27,720 |
0.07736% |
|
case
5 |
4,620 |
0.01289% |
|
case
6 |
138,600 |
0.38681% |
|
case
7 |
415,800 |
1.16042% |
|
case
8 |
92,400 |
0.25787% |
|
case
9 |
138,600 |
0.38681% |
|
case
10 |
2,494,800 |
6.96253% |
|
case
11 |
3,326,400 |
9.28337% |
|
case
12 |
1,247,400 |
3.48126% |
|
case
13 |
9,979,200 |
27.85012% |
|
case
14 |
13,970,880 |
38.99016% |
|
case
15 |
3,991,680 |
11.14005% |
|
Total |
35,831,808 |
100.00000% |
Question 3
For the
denominator, the two remaining people each have 11 floors to choose from for a
total of 11 x 11 = 121. For the
numerator, we have 11P2 = 110. So, the percentage is 110/121 = 90.91%
Question 4
The sample space is:
1: none the same
2: pair
3: 2 pair
4: 3 pair
5: 3 of a kind
6: full house
7: 2 of 3 of a kind (eg. 333777)
8: 4 of a kind
9: 4 of a kind plus pair
Question 5
The basic strategy is to choose the numbers and then the colours for each number
1: none the same – 6 groups of 1
Numbers: 12C6 = 924
Colours: (4C1)(4C1)(4C1)(4C1)(4C1)(4C1) = 46 = 4,096
Total = 924 x 4,096 = 3,784,704
2: pair – 1 group of 2, 4 groups of 1
Numbers: (12C1)(11C4) = 12 x 330
Colours: (4C2) (4C1)(4C1)(4C1)(4C1) = 6 x 44 = 6 x 256
Total = 12 x 330 x 6 x 256 = 6,082,560
3: 2 pair – 2 groups of 2, 2 groups of 1
Numbers: (12C2)(10C2) = 66 x 45
Colours: (4C2)(4C2)(4C1)(4C1) = 62 x 42 = 36 x 16
Total = 66 x 45 x 36 x 16 = 1,710,720
4: 3 pair – 3 groups of 2
Numbers: (12C3) = 220
Colours: (4C2)(4C2)(4C2) = 63 = 216
Total = 220 x 216 = 47,520
5: 3 of kind – 1 group of 3, 3 groups of 1
Numbers: (12C1)(11C3) = 12 x 165
Colours: (4C3)(4C1)(4C1)(4C1) = 44 = 256
Total = 12 x 165 x 256 = 506,880
6: Full house – 1 group of 3, 1 group of 2, 1 group of 1
Numbers: (12C1)(11C1)(10C1) = 12 x 11 x 10
Colours: (4C3)(4C2)(4C1) = 4 x 6 x 4
Total = 12 x 11 x 10 x 4 x 6 x 4 = 126,720
7: 2 of 3 of a kind – 2 groups of 3
Numbers: (12C2) = 66
Colours: (4C3)(4C3) = 42 = 16
Total: 66 x 16 = 1,056
8: 4 of a kind – 1 group of 4, 2 groups of 1
Numbers: (12C1)(11C2) = 12 x 55
Colours: (4C4)(4C1)(4C1) = 1 x 42 = 16
Total = 12 x 55 x 16 = 10,560
9: 4 of a kind plus pair – 1 group of 4, 1 group of 2
Numbers: (12C1)(11C1) = 12 x 11
Colours: (4C4)(4C2) = 1 x 6 = 6
Total = 12 x 11 x 6 = 792
Here are the probabilities:
|
Event |
Number |
Probability |
|
None same |
3,784,704 |
30.84% |
|
Pair |
6,082,560 |
49.57% |
|
2 pair |
1,710,720 |
13.94% |
|
3 pair |
47,520 |
0.39% |
|
3 of kind |
506,880 |
4.13% |
|
full house |
126,720 |
1.03% |
|
2 of 3 of kind |
1,056 |
0.01% |
|
4 of kind |
10,560 |
0.09% |
|
4 of kind plus pair |
792 |
0.01% |
|
Total |
12,271,512 |
100.00% |
Question 6
P(2
threes or 4 red) = P(2 threes) + P(4 red) – P(2 threes and 4 red)
P(2 threes)
= (4C2)(44C4)/48C6 = (6)(135,751)/48C6 = 814,506/48C6 (48C6 = 12,271,512)
P(4 red)
= (12C4)(36C2)/48C6 = (495)(630)/48C6 = 311,850/48C6
For the
last probability, we need the two cases: with the red 3 and without the red 3.
Case 1:
with the red 3
Red three:
1C1 = 1
Other 3:
3C1 = 3
Three other
reds: 11C3 = 165
One other
card: 33C1 = 33
1 x 3 x 165
x 33 = 16,335
Case 2:
without the red 3
Two threes:
3C2 = 3
Four reds:
11C4 = 330
3 x 330 =
990
P(2
threes and 4 red) = (16,335 + 990)/48C6 = 17,325/48C6
P(2
threes or 4 red) = (814,506 + 311,850 – 17,325)/48C6 = 1,109,031/48C6 = 0.0904
= 9.04%
Question 7
The sample
space is {pair, 2 pairs, 3 pairs, full house, pair plus 4 of a kind}.
The total
number of choices for the denominator is 4C2 x 44C4 = 6 x 135,751 = 814,506
since 4 of the cards are sevens and the remaining 44 cards are not sevens.
Now we work
through each event in the sample space:
Pair:
Step 1:
Choose the faces for the remaining 4 cards: 11C4 = 330
Step 2:
Choose the colours for the pair of 7s: 4C2 = 6
Step 3:
Choose the colours for each of the remaining cards: (4C1)4 = 44
= 256
Total = 330
x 6 x 256 = 506,880
Two pairs:
Step 1:
Choose the face for the other pair: 11C1 = 11
Step 2:
Choose the faces for the remaining 2 cards: 10C2 = 45
Step 3: Choose
the colours for the pairs: (4C2)2 = 6x6 = 36
Step 4:
Choose the colours for the remaining 2 cards: (4C1)2 = 4x4 = 16
Total = 11
x 45 x 36 x 16 = 285,120
Three
pairs:
Step 1:
Choose the faces for the other 2 pairs: 11C2 = 55
Step 2: Choose
the colours for each pair: (4C2)3 = 63 = 216
Total = 55
x 216 = 11,880
Full house:
Step 1:
Choose the face for the 3 of a kind: 11C1 = 11
Step 2:
Choose the face for the remaining card: 10C1 = 10
Step 3:
Choose the colours for the pair of 7s: 4C2 = 6
Step 4:
Choose the colours for the 3 of a kind: 4C3 = 4
Step 5:
Choose the colour for the last card: 4C1 = 4
Total = 11
x 10 x 6 x 4 x 4 = 10,560
Pair plus 4
of a kind:
Step 1:
Choose the face for the 4 of a kind: 11C1 = 11
Step 2:
Choose the colours for the pair of 7s: 4C2 = 6
Total = 11
x 6 = 66
Now we can
put together a frequency table:
|
Event |
Frequency |
Percent |
|
Pair |
506,880 |
62.232% |
|
Two
pair |
285,120 |
35.005% |
|
Three
pair |
11,880 |
1.459% |
|
Full
house |
10,560 |
1.296% |
|
Pair/4
of kind |
66 |
0.008% |
|
Total |
814,506 |
100.000% |
Question 8
The total
is 2030 = 1.0737 x 1039 approximately.
For the
numerator:
Choose
stations for the groups of 3: 20C6 = 38,760
Choose
stations for the groups of 2: 14C3 = 364
Choose stations
for the groups of 1: 11C6 = 462
Arrange the
people among the stations: 30!/[(3!)6(2!)3]
= 7.1066 x 1026
Total for
numerator = 38,760 x 364 x 462 x 7.1066 x 1026 = 4.6322 x 1036
Probability
= 4.6322 x 1036/2030 = 0.43%
Question 9
First, we need to choose
the first incorrect letter. For any letter, there are 3 incorrect matches. We
have 3C1 = 3. For the second incorrect letter, we also have 3 incorrect
matches. For example, say we put letter B in envelope A. For envelope B, we would
then choose from letters A, C or D. The total is 3 x 3 = 9 and P(0) = 9/24.
Question 10
The total is 5! = 120. For
X = 5, there is only 1 arrangement: ABCDE. Then P(5) =
1/120
For X = 4, if 4 are
correct, then the fifth match must also be correct. Then P(4)
= 0/120
For X = 3, we choose the 3
correct matches: 5C3 = 10. Then the other 2 matches must be incorrect. Then P(3) = 10/120
For X = 2, we choose the 2
correct matches: 5C2 = 10. We then choose the first incorrect match of which
there are 2 choices. For example, suppose A and B are
matched correctly. Then for envelope C, we can choose letter D or E. As
previously, the last 2 matches must be incorrect. The total is 10 x 2 = 20.
Then P(2) = 20/120.
For X = 1, we choose the
correct match: 5C1 = 5. Next, we choose the first incorrect match. As with
question 9 above, there are 3 incorrect matches. For the second incorrect
match, there are also 3 incorrect matches following the same logic as in
question 9. Again, the last 2 matches must be incorrect. The total is 5 x 3 x 3
= 45. Then P(1) = 45/120.
For X = 0, we can calculate
P(1) + … + P(5) = 76/120. Then, P(0)
= 1 – 76/120 = 44/120. If you can work out this probability
analytically as with the previous probabilities, kudos to you. A hint:
After choosing the first incorrect letter of which there are 4 choices, there
are 2 cases to work through. In the first case, you would have simply switched
two letters (B in envelope A, A in envelope B). For the second case, you would
need to follow the logic of X = 1. You should get 8/120 for case 1 and 36/120
for case 2. Adding the two cases gives 44/120.
From the textbook
Section 3-6
Question 18
Part a: 12P4 = 11,880
Part b: 12C4 = 495
Question 24
Part a: 28 = 256
Part b: 8!/4!4!
= 8C4 = 70
Part c: P(4
boys and 4 girls) = 70/256 = 0.2734 = 27.34%
Question 30
Part a: 220 =
1,048,576
Part b: 20!/10!10! = 20C10 = 184,756
Part c: P(10
boys and 10 girls) = 184,756/1,048,576 = 0.1762 = 17.62%
Part d: There is less than
a 1 in 5 chance of having 10 boys and 10 girls; this does not appear to be a
common occurrence.
Question 34
22P8 = 22!/14!
= 22 x 21 x 20 x 19 x 18 x 17 x 16 x 15 = 12,893,126,400
Question 40
(8C4)(10C4) = 70 x 210 =
14,700