first_draw | second_draw |
|---|---|
R1 | R2 |
R3 | |
Y1 | |
Y2 | |
R2 | R1 |
R3 | |
Y1 | |
Y2 | |
R3 | R1 |
R2 | |
Y1 | |
Y2 | |
Y1 | R1 |
R2 | |
R3 | |
Y2 | |
Y2 | R1 |
R2 | |
R3 | |
Y1 |
Homework Chapter 4
4.1.1-6, 4.1.28
Section 4.1
Q1
\(S = \{1,2,3,4,5,6\} = \{E_1, E_2, E_3, E_4, E_5, E_6\}\)
\(A = \{2\}\)
Q2
\(B = \{2,4,6\}\)
Q3
\(C = \{3,4,5,6\}\)
Q4
\(D = \{1,2,3,4\}\)
Q5
\(E = \{2,4,6\}\)
Q6
\(F = \{\}\)
Q28
5 balls - R1, R2, R3, Y1, Y2
These are the 20 events using a tree like diagram
Section 4.2
[1] "4.2.7-4.2.10, 4.2.27, 4.2.36"4.2.7-4.2.10
7 Events, E7 has twice the probability.
Events 1-6 have probability x, E7 has probability 2x
\[ P(E_1) + P(E_2) + P(E_3) + P(E_4) + P(E_5) + P(E_6) + P(E_7) = 1 \]
\[ x + x + x +x +x +x + 2x = 1 = 8x \]
\[ x = \frac{1}{8} \]
4.2.7
\[ x + x +x = \frac{3}{8} \]
4.2.8
\[ x + x + x + 2x = \frac{5}{8} \]
4.2.9
\[ x + x = \frac{2}{8} \]
4.2.10
All three are not possible - therefore probability = 0
4.2.27
Part a:
Total must some to 1 therefore
\[ 1 = .49 + .21 + .09 + ? \]
Remaining probability = 0.21
Part b:
\[ .49 + .21 + .21 = .91 = 1 - .09 \]
4.2.36
Use tree diagram to show all simple events:
first | second | third | fourth | probability |
|---|---|---|---|---|
John | Bill | Ed | Dave | 1/24 |
Dave | Ed | 1/24 | ||
Ed | Bill | Dave | 1/24 | |
Dave | Bill | 1/24 | ||
Dave | Bill | Ed | 1/24 | |
Ed | Bill | 1/24 | ||
Bill | John | Dave | 1/24 | |
Dave | Ed | 1/24 | ||
Ed | John | Dave | 1/24 | |
Dave | John | 1/24 | ||
Dave | John | Ed | 1/24 | |
Ed | John | 1/24 | ||
Ed | John | Bill | Dave | 1/24 |
Dave | Bill | 1/24 | ||
Bill | John | Dave | 1/24 | |
Dave | John | 1/24 | ||
Dave | John | Bill | 1/24 | |
Bill | John | 1/24 | ||
Dave | John | Ed | 1/24 | |
Ed | Bill | 1/24 | ||
Bill | John | Ed | 1/24 | |
Ed | John | 1/24 | ||
Ed | John | Bill | 1/24 | |
Bill | John | 1/24 |
Part a
first | second | third | fourth | probability |
|---|---|---|---|---|
John | Bill | Ed | Dave | 1/24 |
Dave | Ed | 1/24 | ||
Ed | Bill | Dave | 1/24 | |
Dave | Bill | 1/24 | ||
Dave | Bill | Ed | 1/24 | |
Ed | Bill | 1/24 | ||
Bill | John | Dave | 1/24 | |
Dave | Ed | 1/24 | ||
Ed | John | Dave | 1/24 | |
Dave | John | 1/24 | ||
Dave | John | Ed | 1/24 | |
Ed | John | 1/24 | ||
Ed | John | Bill | Dave | 1/24 |
Dave | Bill | 1/24 | ||
Bill | John | Dave | 1/24 | |
Dave | John | 1/24 | ||
Dave | John | Bill | 1/24 | |
Bill | John | 1/24 | ||
Dave | John | Ed | 1/24 | |
Ed | Bill | 1/24 | ||
Bill | John | Ed | 1/24 | |
Ed | John | 1/24 | ||
Ed | John | Bill | 1/24 | |
Bill | John | 1/24 |
6/24 = 1/4
Part b
first | second | third | fourth | probability |
|---|---|---|---|---|
John | Bill | Ed | Dave | 1/24 |
Dave | Ed | 1/24 | ||
Ed | Bill | Dave | 1/24 | |
Dave | Bill | 1/24 | ||
Dave | Bill | Ed | 1/24 | |
Ed | Bill | 1/24 | ||
Bill | John | Dave | 1/24 | |
Dave | Ed | 1/24 | ||
Ed | John | Dave | 1/24 | |
Dave | John | 1/24 | ||
Dave | John | Ed | 1/24 | |
Ed | John | 1/24 | ||
Ed | John | Bill | Dave | 1/24 |
Dave | Bill | 1/24 | ||
Bill | John | Dave | 1/24 | |
Dave | John | 1/24 | ||
Dave | John | Bill | 1/24 | |
Bill | John | 1/24 | ||
Dave | John | Ed | 1/24 | |
Ed | Bill | 1/24 | ||
Bill | John | Ed | 1/24 | |
Ed | John | 1/24 | ||
Ed | John | Bill | 1/24 | |
Bill | John | 1/24 |
2/24
Part c
first | second | third | fourth | probability |
|---|---|---|---|---|
John | Bill | Ed | Dave | 1/24 |
Dave | Ed | 1/24 | ||
Ed | Bill | Dave | 1/24 | |
Dave | Bill | 1/24 | ||
Dave | Bill | Ed | 1/24 | |
Ed | Bill | 1/24 | ||
Bill | John | Dave | 1/24 | |
Dave | Ed | 1/24 | ||
Ed | John | Dave | 1/24 | |
Dave | John | 1/24 | ||
Dave | John | Ed | 1/24 | |
Ed | John | 1/24 | ||
Ed | John | Bill | Dave | 1/24 |
Dave | Bill | 1/24 | ||
Bill | John | Dave | 1/24 | |
Dave | John | 1/24 | ||
Dave | John | Bill | 1/24 | |
Bill | John | 1/24 | ||
Dave | John | Ed | 1/24 | |
Ed | Bill | 1/24 | ||
Bill | John | Ed | 1/24 | |
Ed | John | 1/24 | ||
Ed | John | Bill | 1/24 | |
Bill | John | 1/24 |
6/24 = 1/4
Section 4.3
4.3.1
\(10 \times 8 = 80\)
4.3.2
\(4 \times 7 \times 3 = 84\)
4.3.5-12
4.3.5: 60
4.3.6: 3628800
4.3.7: 720
4.3.8: 20
4.3.9: 10
4.3.10: 10
4.3.11: 1
4.3.12: 20
4.3.13
\[ P^8_5 = 6720 \]
4.3.16
\[ 4 \times 12 \times 4 = 192 \]
4.3.19
Part A: - 3 stages with n = 52 for each: - \(52 \times 52 \times 52 = 140,608\)
Part B:
- 3 stages with n = 52, 51, and 50, respectively
- \(52 \times 51 \times 50 = 132,600\)
Part C: - 1st stage - pick any card = 52 - 2nd stage - 1 option same card as the previous person - 3rd stage - 1 option - same card as previous
- Total options from part A
- Probability of same card = \(52/140608 = 0.00037\)
Part D
- Probability that all three students pick different cards:
- Part B/Part A = \(132600/140608 = 0.943\)
4.3.21
Part A: order does not matter - \(C^{52}_5 = \frac{52!}{(52 - 5)!5!} = 2,598,960\)
Part B: Only 4 different combinations
Part C: 4/2598960 = 0.00000154
4.3.24
Part A
- \(C^5_2 = 30\)
Part B:
- 1/30
4.3.24
4.3.26
Part A: \(4 \times 3 \times 2 \times 1 = 24\)
Part B: If Dave wins sprint there are \(1 \times 3 \times 2 \times 1 = 6\) total ways he can win. Probability = 6/24
Part C: \(1 \times 1 \times 2 \times 1 = 2\), Probability = 2/24
Part D: There are 3 options for first (any but Ed), 2 options for second (not and and not first), 1 option for 3rd left and Ed for fourth. Prob = 6/24
4.3.28
- Step 1 - how many ways can 5 problems go on the exam from 10 questions?
- \(C^{10}_5 = 252\)
- Step 2 - howmany ways can you choose 5 problems from 6 questions?
- \(C^6_5 = 6\)
Probability of all 5 being questions she knows is 6/252 = 0.024.
4.3.29
Total ways to draw the shapes: \(12! = 479,001,600\)
How many ways to group them when drawing them? \(12 \times 2 \times 1 \times 9 \times 2 \times 1 \times 6 \times 2 \times 1 \times 3 \times 2 \times 1 = 31,104\)
Probability = \(31,104/479,001,600 = 0.00006494 = \frac{4!(3!)^4}{12!}\)
Section 4.4
4.4.1
- a: \(P(A\^c) = P(\\{E_2, E_4, E_5\\}) = .6\)
- b: \(P(A \cap B) = P(\{E_1\}) = .2\)
- c: \(P(B \cap C) = P(\{E_4\}) = .2\)
- d: \(P(A \cup B) = P(\{E_1, E_2, E_3, E_4, E_5\}) = 1\)
- e: \(P(B|C) = \frac{P(B\cap C)}{P(C)} = \frac{.2}{.4} = .5\)
- f: \(P(A|B) = \frac{P(A\cap B)}{P(B)} = \frac{.2}{.8} = .25\)
- g: \(P(A \cup B \cup C) = P(\{E_1, E_2, E_3, E_4, E_5\}) = 1\)
- h: \(P(\left(A \cap B \right)^ c) = 1 - .2 = .8\)
4.4.7 - 4.4.10
Q7
\[ P(A|B) = .1 = \frac{P(A\cap B)}{P(B)} \]
\(P(A\cap B) = .05\)
Q8
Yes
Q9
No
Q10
No
4.4.14
- a: Yes, \(.3 \times .4 = .12\)
- b: 0 - they are mutually exclusive - use the addition rule
- c: \(P(A) = .3\)
- d: 0
4.4.27
\(1 - 1/1000 \times 1/1000 = 1 - 1/1000000 = .999999\)
4.4.29
- a: \(.7 \times .6 = .42\)
- b: Yes - probability of Mocha is the same regardless of where she goes
- c: 30%
- d: \(P(A \cup B) = P(A) + P(B) - P(A\cap B) = .7 + .6 - .42 = .88\)
4.4.32
\(P(A) = .95\), \(P(B) = .98\), \(P(A\cap B) = .94\)
- a: \(P(A \cup B) = P(A) + P(B) - P(A \cap B) = .99\)
- b: \(1 - P(A \cup B) = .01\)
4.4.35
- a: \(P(A) = 154/256 = 0.60\)
- b: \(P(G) = 155/256 = 0.61\)
- c: \(P(A \cap G) = 88/256 = 0.34\)
- d: \(P(G|A) = 88/154 = 0.57\)
- e: \(P(G|B) = 44/67 = 0.66\)
- f: \(P(G|C) = 23/35 = 0.66\)
- g: \(P(C|P) = 12/101 = 0.12\)
- h: \(P(B^c) = 1 - 67/256 = 0.74\)
Section 4.5
4.5.1
\[ P(A) = P(A|S_1)P(S_1) + P(A|S_2)P(S_2) = .7 * .2 +.3 * .3 = 0.23 \]
4.5.2
\[ P(S_1|A) = \frac{P(A|S_1)P(S_1)}{P(A)} = \frac{.2 * .7}{.23} = 0.61 \]
4.5.3
\[ P(S_2|A) = \frac{P(A|S_2)P(S_2)}{P(A)} = \frac{.3 * .3}{.23} = 0.39 \]
4.5.9
Event D - defective item
Event M - worker reads manual
\(P(D|M) = .01\)
\(P(D|M^c) = .03\)
\(P(M) = .9\)
\[ P(D) = P(D|M)P(M) + P(D|M^c)P(M^c) . = .01*.9 + .03*(1-.9) = 0.012 \]
4.5.15
Part a
- \(P(D) = .1\)
- \(P(D^c) = .9\)
- \(P(N|D^c) = .85/.9 = 0.94\)
- \(P(N|D) = .02/.1 = .20\)
Part b
\[ P(D|N) = \frac{P(N|D)P(D)}{P(N)} = \frac{P(N|D)P(D)}{P(N|D)P(D) + P(N|D^c)P(D^c)} = \frac{.2*.1}{.2*.1 + .94*.9} = 0.023 \]
Part C
- note that answers are different at 5th decimal point because of rounding in part A
\[ P(D|N) = P(D\cap N)/P(N) = .02/.87 = 0.023 \]
Part D
\[ P(P|D^c) = .05/(.05 + .85) = 0.056 \]
Part E
\[ P(N|D) = .2 \]
Part f
The false negative rate is a little high - may miss cases.