Complete Concept Guide (100% Curriculum Coverage)
1. Sample Space & Types of Events
The set of all possible outcomes of a random experiment is the Sample Space ($S$):
• Tossing 2 coins: $S = \{HH, HT, TH, TT\}$.
• Mutually Exclusive Events: Two events $A$ and $B$ that cannot occur simultaneously: $\mathbf{A \cap B = \emptyset \implies P(A \cap B) = 0}$.
• Exhaustive Events: Events whose union covers the entire sample space: $\mathbf{A \cup B = S}$.
2. Axiomatic Definition of Probability
For any event $E$ of sample space $S = \{\omega_1, \omega_2, \dots, \omega_n\}$:
(1) $0 \le P(\omega_i) \le 1$ for all outcomes.
(2) $\sum_{i=1}^{n} P(\omega_i) = 1$.
(3) For any event $E$, $P(E) = \sum_{\omega \in E} P(\omega)$.
3. Probability Addition Theorems
For any two events $A$ and $B$: $$\mathbf{P(A \cup B) = P(A) + P(B) - P(A \cap B)}$$ If $A$ and $B$ are mutually exclusive: $$\mathbf{P(A \cup B) = P(A) + P(B)}$$ Complement: $\mathbf{P(A') = 1 - P(A)}$.
Check Your Understanding (Diagnostic Practice Questions)
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1
A coin is tossed three times. Write the sample space and find the probability of getting at least two heads.
Reveal Answer & Explanation
Answer: $S = \{HHH, HHT, HTH, THH, HTT, THT, TTH, TTT\}$ ($n(S) = 8$). Event of at least 2 heads $= \{HHH, HHT, HTH, THH\}$ (4 outcomes). Probability $= \frac{4}{8} = \frac{1}{2}$.
Sample space has 8 outcomes; P = 1/2.
2
If $P(A) = \frac{3}{5}$ and $P(B) = \frac{1}{5}$, find $P(A \text{ or } B)$ if $A$ and $B$ are mutually exclusive events.
Reveal Answer & Explanation
Answer: For mutually exclusive events, $P(A \cap B) = 0$. $P(A \cup B) = P(A) + P(B) = \frac{3}{5} + \frac{1}{5} = \frac{4}{5}$.
4/5.
3
If $P(A) = 0.5$, $P(B) = 0.3$, and $P(A \cap B) = 0.2$, find: (i) $P(A \cup B)$, (ii) $P(A')$.
Reveal Answer & Explanation
Answer: (i) $P(A \cup B) = P(A) + P(B) - P(A \cap B) = 0.5 + 0.3 - 0.2 = 0.6$. (ii) $P(A') = 1 - P(A) = 1 - 0.5 = 0.5$.
(i) 0.6; (ii) 0.5.
4
Two dice are thrown simultaneously. What is the probability of getting a sum greater than 9?
Reveal Answer & Explanation
Answer: Total outcomes $= 36$. Favorable outcomes with sum $> 9$ (sums 10, 11, 12): $(4,6), (5,5), (6,4), (5,6), (6,5), (6,6)$ (6 pairs). $P = \frac{6}{36} = \frac{1}{6}$.
1/6.
5
Explain why $P(A \cup B) = P(A) + P(B)$ is true only for mutually exclusive events.
Reveal Answer & Explanation
Answer: Because for mutually exclusive events, the intersection $A \cap B = \emptyset$, so $P(A \cap B) = 0$. When events overlap, the intersection probability must be subtracted to prevent double-counting shared elements.
Shared intersection is empty; otherwise double-counting occurs.
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