Unit 14: Probability (सम्भाव्यता) Formulae & Solutions
Solutions by Important Edu Notes Team
1. Probability Formulas & Rules
| Concept | Rule / Formula | Description (Nepali) |
|---|---|---|
| Basic Probability | $$P(E) = \frac{n(E)}{n(S)}$$ | सम्भाव्यता = अनुकूल परिणामहरूको सङ्ख्या / कुल परिणामहरूको सङ्ख्या |
| Mutually Exclusive Events | $$P(A \cup B) = P(A) + P(B)$$ | यदि A र B परस्पर निषेधक छन् भने, A वा B को सम्भाव्यता जोडिन्छ। |
| Non-Mutually Exclusive Events | $$P(A \cup B) = P(A) + P(B) – P(A \cap B)$$ | यदि A र B परस्पर निषेधक होइनन् भने, A र B दुवै एकैचोटि हुन सक्छन्। |
| Independent Events | $$P(A \cap B) = P(A) \times P(B)$$ | यदि A र B स्वतन्त्र घटनाहरू छन् भने, A र B दुवै एकैसाथ हुने सम्भाव्यता गुणन हुन्छ। |
| Complementary Events | $$P(A’) = 1 – P(A)$$ | A नहुने सम्भाव्यता 1 मा A हुने सम्भाव्यता घटाउँदा पाइन्छ। |
Key Terms (मुख्य शब्दावली):
- Sample Space (प्रतिदर्श समूह): All possible outcomes of an experiment.
- Event (घटना): A subset of sample space.
- Mutually Exclusive (परस्पर निषेधक): Events that cannot occur at the same time.
- Independent Events (स्वतन्त्र घटनाहरू): Occurrence of one does not affect the other.
2. Exercise 14.1 – Q1: Mutually Exclusive Events
Given: Sample space $S = \{H, T\}$, $A = \{H\}$, $B = \{T\}$
Since $A \cap B = \phi$, there are no common outcomes. Both cannot happen at the same time.
Given: $S = \{1,2,3,4,5,6\}$, $P = \{2,4,6\}$, $Q = \{1,3,5\}$
Given: Face cards = King, Queen, Jack of all suits. Spade cards include King, Queen, Jack of Spades.
There are common cards (face cards that are also spades).
Given: A card cannot be both ’10’ and ‘Ace’ at the same time.
A single ball drawn cannot be both green and blue.
3. Exercise 14.1 – Q2: Basic Probability
Sample Space: $S = \{HH, HT, TH, TT\}$, $n(S) = 4$
Event E (at least one head): $E = \{HH, HT, TH\}$, $n(E) = 3$
Sample Space: $S = \{1,2,3,4,5,6\}$, $n(S) = 6$
Prime numbers: $\{2,3,5\}$, $n(E) = 3$
Total cards: $n(S) = 52$
Face cards: King, Queen, Jack of 4 suits = $3 \times 4 = 12$
Total months: $n(S) = 12$
Months with 30 days: April, June, September, November → $n(E) = 4$
Total balls: $4+7+5 = 16$, $n(S) = 16$
White balls: $n(E) = 4$
4. Exercise 14.1 – Q3: Probability of “OR” Events
$n(S)=18$, $n(R)=6$, $n(B)=7$. Mutually exclusive.
$n(S)=8$, $A=\{HHH\}$, $B=\{TTT\}$.
Prime numbers = $\{2,3,5\}$, $n(P)=3$, $F=\{4\}$, $n(F)=1$.
$n(S)=52$, $n(10)=4$, $n(Ace)=4$.
Not mutually exclusive: Some face cards are spades.
$P(F)=\frac{12}{52}$, $P(Sp)=\frac{13}{52}$, $P(F \cap Sp)=\frac{3}{52}$
5. Exercise 14.1 – Q4: Word Problems
Total letters = 11, M’s = 2, T’s = 2.
Total letters = 10, S’s = 3, T’s = 3.
Total letters = 12, O’s = 3, D’s = 3.
$n(S)=15$, $n(E)=8$, $n(M)=9$, $n(E \cap M)=4$.
6. Exercise 14.2: Compound Events
Independent events:
$P(T) = \frac{1}{2}$, $P(3) = \frac{1}{6}$
Total balls = 10. With replacement → independent events.
(a) Both same color:
(b) Both different color:
(c) At least one red or black:
Without replacement → dependent events.
(a) Both same color:
(b) Both different color:
(c) At least one red or black:
Total = 15. Without replacement.
$P(RR) = \frac{7}{15} \times \frac{6}{14} = \frac{42}{210} = \frac{1}{5}$
$P(YY) = \frac{8}{15} \times \frac{7}{14} = \frac{56}{210} = \frac{4}{15}$
(a) Both are aces:
(b) One ace and one king (order matters):
All possible sequences of drawing all 3 marbles:
RGB, RBG, GRB, GBR, BRG, BGR → 6 equally likely sequences.
7. Exercise 14.3: Tree Diagrams
Sample Space (8 outcomes): HHH, HHT, HTH, HTT, THH, THT, TTH, TTT
HT
Full tree has 8 terminal branches.
(a) All three tails: $P(TTT) = \frac{1}{8}$
(b) At least two heads: Outcomes = {HHH, HHT, HTH, THH} → $P = \frac{4}{8} = \frac{1}{2}$
(c) Three tails: Same as (a) → $\frac{1}{8}$
Sample Space (6 outcomes): RH, RT, BH, BT, BrH, BrT
(a) Red on spinner and Head on coin:
(b) Brown on spinner and (Tail or Head) on coin:
Sample Space (12 outcomes): H1, H2,…, T6
(a) Head and even number (2,4,6):
(b) Tail and square number (1,4):
(a) Red card and Head:
(b) Black card and Tail:
Total marbles = 12
(i) With replacement: Probabilities remain constant.
Example: $P(RRR) = (\frac{7}{12})^3 = \frac{343}{1728}$
(ii) Without replacement: Probabilities change.
Example: $P(RRR) = \frac{7}{12} \times \frac{6}{11} \times \frac{5}{10} = \frac{210}{1320} = \frac{7}{44}$
Tree diagram would show all 8 branches (RRR, RRG, RGR, RGG, GRR, GRG, GGR, GGG) with respective probabilities.