Probability
Exercise 14.2 Solutions
Probability Basics:
1. Total Probability Axiom: $\sum P(\omega_i) = 1$
2. Range: $0 \le P(\omega_i) \le 1$ for all events.
3. Probability of Event E: $P(E) = \frac{\text{No. of outcomes in E}}{\text{Total outcomes in S}}$
1. Total Probability Axiom: $\sum P(\omega_i) = 1$
2. Range: $0 \le P(\omega_i) \le 1$ for all events.
3. Probability of Event E: $P(E) = \frac{\text{No. of outcomes in E}}{\text{Total outcomes in S}}$
Q1
1. Which assignments of probability are valid for S = {$\omega_1, \dots, \omega_7$}?
| Case | Values Check ($0 \le p \le 1$) | Sum Check ($\sum p_i = 1$) | Valid? |
|---|---|---|---|
| (a) | All values are between 0 and 1. | $0.1+0.01+0.05+0.03+0.01+0.2+0.6 = 1.00$ | Yes |
| (b) | All $1/7$. OK. | $7 \times (1/7) = 1$. | Yes |
| (c) | All positive. OK. | $0.1+0.2+0.3+0.4+0.5+0.6+0.7 = 2.8 \neq 1$. | No |
| (d) | Contains negative values (-0.1, -0.2). | Not checked (Fails condition 1). | No |
| (e) | $15/14 > 1$. Invalid. | Not checked. | No |
Q2
2. Coin tossed twice. Probability of at least one tail?
Sample Space
$S = \{HH, HT, TH, TT\}$. Total = 4.
Event E
At least one tail: $\{HT, TH, TT\}$. Count = 3.
$P(E) = \frac{3}{4}$
Q3
3. Die thrown. Find probabilities. $S = \{1, 2, 3, 4, 5, 6\}$ (Total 6).
(i) Prime
$\{2, 3, 5\}$ (3 outcomes). $P = 3/6 = 1/2$.
(ii) $\ge 3$
$\{3, 4, 5, 6\}$ (4 outcomes). $P = 4/6 = 2/3$.
(iii) $\le 1$
$\{1\}$ (1 outcome). $P = 1/6$.
(iv) > 6
$\phi$ (0 outcomes). $P = 0$.
(v) < 6
$\{1, 2, 3, 4, 5\}$ (5 outcomes). $P = 5/6$.
Q4
4. Card from pack of 52.
(a) Points
Sample space contains 52 points.
(b) Ace Spade
Only 1 Ace of Spades. $P = 1/52$.
(c)(i) Ace
4 Aces. $P = 4/52 = 1/13$.
(c)(ii) Black
26 Black Cards (Spades + Clubs). $P = 26/52 = 1/2$.
Q5
5. Coin marked 1, 6 and normal Die. Find P(Sum=3) and P(Sum=12).
Sample Space
Coin $\{1, 6\}$. Die $\{1, 2, 3, 4, 5, 6\}$. Total outcomes $2 \times 6 = 12$.
$S = \{(1,1), (1,2), (1,3), (1,4), (1,5), (1,6), (6,1), (6,2), (6,3), (6,4), (6,5), (6,6)\}$
$S = \{(1,1), (1,2), (1,3), (1,4), (1,5), (1,6), (6,1), (6,2), (6,3), (6,4), (6,5), (6,6)\}$
(i) Sum 3
Only outcome is $(1, 2)$. Count = 1. $P = 1/12$.
(ii) Sum 12
Only outcome is $(6, 6)$. Count = 1. $P = 1/12$.
Q6
6. Council: 4 men, 6 women. One selected. Probability it’s a woman?
Total
$4 + 6 = 10$ members.
Women
6 members.
$P(W) = \frac{6}{10} = \frac{3}{5}$
Q7
7. 4 Coin Tosses. Win Re 1 (Head), Lose Rs 1.50 (Tail). Amounts and Probabilities.
Total
$2^4 = 16$ outcomes.
| Event | Outcomes | Calculation | Amount | Prob. |
|---|---|---|---|---|
| 4 Heads | 1 (HHHH) | $4 \times 1$ | Rs 4.00 | 1/16 |
| 3H, 1T | 4 (HHHT, etc) | $3(1) – 1.5$ | Rs 1.50 | 4/16 = 1/4 |
| 2H, 2T | 6 (HHTT, etc) | $2(1) – 2(1.5)$ | – Rs 1.00 (Loss) | 6/16 = 3/8 |
| 1H, 3T | 4 (HTTT, etc) | $1(1) – 3(1.5)$ | – Rs 3.50 (Loss) | 4/16 = 1/4 |
| 4 Tails | 1 (TTTT) | $- 4(1.5)$ | – Rs 6.00 (Loss) | 1/16 |
Q8
8. Three coins tossed once. (Total Outcomes = 8)
Outcomes
HHH (1), HHT, HTH, THH (3), HTT, THT, TTH (3), TTT (1).
(i) 3 Heads
1 outcome. $P = 1/8$.
(ii) 2 Heads
3 outcomes. $P = 3/8$.
(iii) $\ge$ 2 Heads
3H (1) + 2H (3) = 4 outcomes. $P = 4/8 = 1/2$.
(iv) $\le$ 2 Heads
All except 3H. $8-1=7$. $P = 7/8$.
(v) No Head
TTT (1 outcome). $P = 1/8$.
(vi) 3 Tails
1 outcome. $P = 1/8$.
(vii) Exactly 2 Tails
Means exactly 1 Head. 3 outcomes. $P = 3/8$.
(viii) No Tail
HHH (1 outcome). $P = 1/8$.
(ix) $\le$ 2 Tails
All except 3 Tails. $8-1=7$. $P = 7/8$.
Q9-Q10
9. If $P(A) = 2/11$, find $P(\text{not } A)$.
10. Word ‘ASSASSINATION’. Prob of Vowel/Consonant.
Formula
$P(\text{not } A) = 1 – P(A)$.
$1 – \frac{2}{11} = \frac{9}{11}$
10. Word ‘ASSASSINATION’. Prob of Vowel/Consonant.
Total
Total letters = 13.
Vowels
A, A, A, I, A, I, O. (Wait: A,S,S,A,S,S,I,N,A,T,I,O,N).
Vowels: A, A, A, I, I, O. Count = 6.
Vowels: A, A, A, I, I, O. Count = 6.
Consonants
S, S, S, S, N, T, N. Count = 7.
(i) Vowel: $6/13$. (ii) Consonant: $7/13$.
Probability
Exercise 14.2 Solutions (Q11 – Q21)
Key Formulas:
Addition Theorem: $P(A \cup B) = P(A) + P(B) – P(A \cap B)$
Mutually Exclusive: $P(A \cap B) = 0 \Rightarrow P(A \cup B) = P(A) + P(B)$
De Morgan’s: $P(A’ \cap B’) = P((A \cup B)’) = 1 – P(A \cup B)$
Addition Theorem: $P(A \cup B) = P(A) + P(B) – P(A \cap B)$
Mutually Exclusive: $P(A \cap B) = 0 \Rightarrow P(A \cup B) = P(A) + P(B)$
De Morgan’s: $P(A’ \cap B’) = P((A \cup B)’) = 1 – P(A \cup B)$
Q11
11. Lottery: Choose 6 distinct numbers from 1 to 20. Match fixed 6 to win. Find prob.
Combinations
Total ways to choose 6 numbers from 20 is $^{20}C_6$.
Calc Total
$^{20}C_6 = \frac{20 \times 19 \times 18 \times 17 \times 16 \times 15}{6 \times 5 \times 4 \times 3 \times 2 \times 1} = 38760$.
Event E
Winning combination is unique (order doesn’t matter). $n(E) = 1$.
Probability = $\frac{1}{38760}$
Q12
12. Check consistency of defined probabilities.
(i) $P(A)=0.5, P(B)=0.7, P(A \cap B)=0.6$
Check
$P(A \cap B)$ cannot be greater than $P(A)$. Here $0.6 > 0.5$.
Not Consistently Defined
(ii) $P(A)=0.5, P(B)=0.4, P(A \cup B)=0.8$
Formula
$P(A \cup B) = P(A) + P(B) – P(A \cap B)$.
Solve
$0.8 = 0.5 + 0.4 – P(A \cap B) \Rightarrow P(A \cap B) = 0.9 – 0.8 = 0.1$.
Check
$0 \le 0.1 \le P(A)$ and $P(B)$. This is valid.
Consistently Defined
Q13
13. Fill in the blanks. Formula: $P(A \cup B) = P(A) + P(B) – P(A \cap B)$.
| Case | P(A) | P(B) | P(A $\cap$ B) | P(A $\cup$ B) Calculation |
|---|---|---|---|---|
| (i) | 1/3 | 1/5 | 1/15 | $\frac{1}{3} + \frac{1}{5} – \frac{1}{15} = \frac{5+3-1}{15} = \frac{7}{15}$ |
| (ii) | 0.35 | 0.5 | 0.25 | $0.6 = 0.35 + P(B) – 0.25 \Rightarrow P(B) = 0.5$ |
| (iii) | 0.5 | 0.35 | 0.15 | $0.7 = 0.5 + 0.35 – x \Rightarrow x = 0.85 – 0.7 = 0.15$ |
Q14
14. $P(A)=3/5, P(B)=1/5$. Mutually exclusive. Find $P(A \text{ or } B)$.
Condition
Mutually exclusive $\Rightarrow P(A \cap B) = 0$.
Formula
$P(A \cup B) = P(A) + P(B) = \frac{3}{5} + \frac{1}{5} = \frac{4}{5}$.
4/5
Q15
15. $P(E)=1/4, P(F)=1/2, P(E \cap F)=1/8$.
(i) E or F
$P(E \cup F) = \frac{1}{4} + \frac{1}{2} – \frac{1}{8} = \frac{2+4-1}{8} = \frac{5}{8}$.
(ii) not E & not F
$P(E’ \cap F’) = P((E \cup F)’) = 1 – P(E \cup F) = 1 – \frac{5}{8} = \frac{3}{8}$.
Q16
16. $P(\text{not } E \text{ or not } F) = 0.25$. Are E and F mutually exclusive?
Logic
$P(E’ \cup F’) = P((E \cap F)’) = 1 – P(E \cap F)$.
Calc
$0.25 = 1 – P(E \cap F) \Rightarrow P(E \cap F) = 0.75$.
Conclusion
Since $P(E \cap F) = 0.75 \neq 0$, they are NOT mutually exclusive.
No
Q17
17. $P(A)=0.42, P(B)=0.48, P(A \cap B)=0.16$.
(i) not A
$1 – 0.42 = 0.58$.
(ii) not B
$1 – 0.48 = 0.52$.
(iii) A or B
$0.42 + 0.48 – 0.16 = 0.90 – 0.16 = 0.74$.
Q18
18. 40% Maths, 30% Biology, 10% Both. Prob(Maths or Biology)?
Values
$P(M)=0.40, P(B)=0.30, P(M \cap B)=0.10$.
Formula
$P(M \cup B) = P(M) + P(B) – P(M \cap B)$.
Calc
$0.40 + 0.30 – 0.10 = 0.60$.
0.60
Q19
19. Pass Exam 1: 0.8, Exam 2: 0.7, At least one: 0.95. Pass both?
Values
$P(A)=0.8, P(B)=0.7, P(A \cup B)=0.95$.
Intersection
$P(A \cap B) = P(A) + P(B) – P(A \cup B)$.
Calc
$0.8 + 0.7 – 0.95 = 1.5 – 0.95 = 0.55$.
0.55
Q20
20. Pass Both (E and H) = 0.5. Pass Neither = 0.1. Pass English = 0.75. Find Pass Hindi.
Given
$P(E \cap H) = 0.5, P(E’ \cap H’) = 0.1, P(E) = 0.75$.
Step 1
$P(E’ \cap H’) = 1 – P(E \cup H) \Rightarrow 0.1 = 1 – P(E \cup H) \Rightarrow P(E \cup H) = 0.9$.
Step 2
$P(E \cup H) = P(E) + P(H) – P(E \cap H)$.
$0.9 = 0.75 + P(H) – 0.5$.
$0.9 = 0.75 + P(H) – 0.5$.
Solve
$0.9 = 0.25 + P(H) \Rightarrow P(H) = 0.65$.
0.65
Q21
21. 60 Students. 30 NCC, 32 NSS, 24 Both.
Values
$n(A)=30, n(B)=32, n(A \cap B)=24, Total=60$.
(i) NCC or NSS
$n(A \cup B) = 30 + 32 – 24 = 38$.
$P(A \cup B) = 38/60 = 19/30$.
$P(A \cup B) = 38/60 = 19/30$.
(ii) Neither
$n((A \cup B)’) = 60 – 38 = 22$.
$P = 22/60 = 11/30$.
$P = 22/60 = 11/30$.
(iii) NSS but not NCC
$n(B – A) = n(B) – n(A \cap B) = 32 – 24 = 8$.
$P = 8/60 = 2/15$.
$P = 8/60 = 2/15$.