Mathematics: Class 11 — Chapter 14: Probability

An event is mathematically defined as any subset of the sample space $S$. If the actual outcome $\omega$ of a random experiment belongs to the subset $E$ (i.e., $\omega \in E$), we say that the event $E$ has occurred.

• Sample Space ($S$): The set of all possible outcomes. For example, tossing a coin twice yields $S = \{HH, HT, TH, TT\}$.
• Event ($E$): A subset $E \subseteq S$. For example, the event of getting exactly one head is $E = \{HT, TH\}$.

Events can be classified based on the number of sample points (elements) they contain:

• Complementary Event ('not A'): For any event $A$, the complementary event $A'$ (or $S - A$) consists of all outcomes in $S that are not in $A$. Mathematically, $A' = \{\omega \in S : \omega \notin A\}$.
• The Event 'A or B' (Union): Denoted by $A \cup B$, this event occurs if either $A$ occurs, $B$ occurs, or both occur. Mathematically, $A \cup B = \{\omega : \omega \in A \text{ or } \omega \in B\}$.
• The Event 'A and B' (Intersection): Denoted by $A \cap B$, this event occurs only when both $A$ and $B$ occur simultaneously. Mathematically, $A \cap B = \{\omega : \omega \in A \text{ and } \omega \in B\}$.
• The Event 'A but not B' (Difference): Denoted by $A - B$ or $A \cap B'$, this event occurs when $A$ occurs but $B$ does not. Mathematically, $A - B = \{\omega : \omega \in A \text{ and } \omega \notin B\}$.

• Mutually Exclusive Events: Two events $A$ and $B$ are called mutually exclusive if they cannot occur simultaneously. In set notation, $A \cap B = \phi$. For example, in rolling a die, getting an odd number and getting an even number are mutually exclusive.
• Exhaustive Events: A collection of events $E_1, E_2, \dots, E_n$ is called exhaustive if their union equals the entire sample space $S$. That is, $\bigcup_{i=1}^n E_i = S$. At least one of these events must occur whenever the experiment is performed.
• Mutually Exclusive and Exhaustive Events: If a set of events $E_1, E_2, \dots, E_n$ satisfies both conditions (i.e., $E_i \cap E_j = \phi$ for all $i \neq j$, and $\bigcup_{i=1}^n E_i = S$), they partition the sample space.

Mathematical Problem Statement: A coin is tossed three times. Let $A$ be the event 'No head appears', $B$ be the event 'Exactly one head appears', and $C$ be the event 'At least two heads appear'. Do they form a set of mutually exclusive and exhaustive events?

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Step-by-Step Algorithmic Solution Trace:

1. Write down the sample space: $S = \{HHH, HHT, HTH, THH, HTT, THT, TTH, TTT\}$.
2. Identify the subsets for each event:
   • $A = \{TTT\}$
   • $B = \{HTT, THT, TTH\}$
   • $C = \{HHT, HTH, THH, HHH\}$
3. Check the union of the events: $$A \cup B \cup C = \{TTT, HTT, THT, TTH, HHT, HTH, THH, HHH\} = S$$ Since the union equals $S$, the events are exhaustive.
4. Check pairwise intersections:
   • $A \cap B = \phi$
   • $A \cap C = \phi$
   • $B \cap C = \phi$
   Since all pairwise intersections are empty, the events are mutually exclusive.

Final Evaluation Output: Yes, the events $A$, $B$, and $C$ form a set of mutually exclusive and exhaustive events.

For any sample space $S$ and events $E, F \subseteq S$:

1. Axiom 1 (Non-negativity): For any event $E$, the probability is non-negative: $$P(E) \ge 0$$
2. Axiom 2 (Certainty): The probability of the entire sample space (sure event) is 1: $$P(S) = 1$$
3. Axiom 3 (Additivity): If $E$ and $F$ are mutually exclusive events ($E \cap F = \phi$), then: $$P(E \cup F) = P(E) + P(F)$$

Immediate Consequences:

• The probability of the impossible event is zero: $P(\phi) = 0$.
• For a finite sample space $S = \{\omega_1, \omega_2, \dots, \omega_n\}$, the sum of probabilities of all elementary outcomes is 1: $\sum_{i=1}^n P(\omega_i) = 1$.
• For any event $A$, its probability is the sum of the probabilities of its constituent outcomes: $P(A) = \sum_{\omega_i \in A} P(\omega_i)$.

When all outcomes in a finite sample space $S$ of size $n$ have an equal chance of occurring, they are called equally likely outcomes. Under this assumption:

• The probability of each elementary outcome $\omega_i$ is: $$P(\omega_i) = \frac{1}{n}$$
• For any event $E$ containing $m$ outcomes: $$P(E) = \frac{m}{n} = \frac{\text{Number of outcomes favourable to } E}{\text{Total possible outcomes}}$$

Using the axioms, we can derive key formulas for calculating probabilities of combined events:

• Probability of 'A or B' (General Addition Rule): For any two events $A$ and $B$: $$P(A \cup B) = P(A) + P(B) - P(A \cap B)$$
• Probability of 'not A' (Complement Rule): For any event $A$: $$P(A') = 1 - P(A)$$

Mathematical Problem Statement: Two students, Anil and Ashima, appear in an examination. The probability that Anil will qualify is 0.05 and that Ashima will qualify is 0.10. The probability that both will qualify is 0.02. Find the probability that: (a) Both will not qualify, (b) At least one of them will not qualify, (c) Only one of them will qualify.

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Step-by-Step Algorithmic Solution Trace:

1. Define events: Let $E$ be the event 'Anil qualifies' and $F$ be the event 'Ashima qualifies'. Given: $P(E) = 0.05$, $P(F) = 0.10$, and $P(E \cap F) = 0.02$.
2. Part (a): Both will not qualify ($E' \cap F'$)
   • By De Morgan's Law, $E' \cap F' = (E \cup F)'$.
   • First, find $P(E \cup F) = P(E) + P(F) - P(E \cap F) = 0.05 + 0.10 - 0.02 = 0.13$.
   • Then, $P(E' \cap F') = 1 - P(E \cup F) = 1 - 0.13 = 0.87$.
3. Part (b): At least one of them will not qualify
   • This is the complement of 'both qualify': $1 - P(E \cap F) = 1 - 0.02 = 0.98$.
4. Part (c): Only one of them will qualify
   • This event is $(E \cap F') \cup (E' \cap F)$. Since these are mutually exclusive: $$P(\text{only one}) = P(E \cap F') + P(E' \cap F)$$ $$= [P(E) - P(E \cap F)] + [P(F) - P(E \cap F)]$$ $$= [0.05 - 0.02] + [0.10 - 0.02] = 0.03 + 0.08 = 0.11$$

Final Evaluation Output: (a) 0.87, (b) 0.98, (c) 0.11