Random Experiment & Events — JEE Main Maths PYQ MCQs with Solutions

Solution: When a coin is tossed, there are 2 possible outcomes (H or T). For 4 tosses, the total number of outcomes = $2^4 = 16$. Each outcome is a sequence like HHHH, HHHT, HHTH, ..., TTTT.

Solution: Each die has 6 faces. When two dice are rolled, the first die can show any of 6 outcomes and the second die can show any of 6 outcomes. Total outcomes = $6 \times 6 = 36$.

Solution: Without replacement, order matters (we care about which card is drawn first, second, third). The number of ordered outcomes is $52 \times 51 \times 50 = \frac{52!}{49!} = P(52,3)$.

Solution: An event must be a subset of the sample space. $\{2, 4, 6\} \subseteq S$ is a valid event. The other options contain elements not in $S$ or are supersets of $S$.

Solution: Two events are mutually exclusive (or disjoint) if they cannot occur simultaneously, i.e., they have no outcomes in common. This means $A \cap B = \emptyset$ (their intersection is empty).

Solution: If $A \subseteq B$ (A is a subset of B), then every element of $A$ is also in $B$. Therefore, the intersection $A \cap B$ contains exactly the elements that are in both, which is all elements of $A$. Hence, $A \cap B = A$.

Solution: For any event $A$ in sample space $S$, the complement $A^c$ contains all outcomes not in $A$. Therefore, $A \cup A^c$ contains all outcomes in $A$ plus all outcomes not in $A$, which is the entire sample space $S$.

Solution: De Morgan's first law: $(A \cup B)^c = A^c \cap B^c$. The complement of a union is the intersection of the complements. This means: not (A or B) = (not A) and (not B).

Solution: De Morgan's second law: $(A \cap B)^c = A^c \cup B^c$. The complement of an intersection is the union of the complements. This means: not (A and B) = (not A) or (not B).

Solution: If $A \cap B = \emptyset$ (A and B are disjoint) and $A \cup B = S$ (together they cover the entire sample space), then $B$ contains exactly those outcomes not in $A$. This is the definition of the complement of $A$, denoted $B = A^c$.

Solution: Two events are exhaustive if together they cover the entire sample space, i.e., $A \cup B = S$. This means at least one of $A$ or $B$ must occur in every outcome.

Solution: If $A$, $B$, $C$ are mutually exclusive and exhaustive: (1) Mutually exclusive means no two events have common outcomes: $A \cap B = \emptyset$, $B \cap C = \emptyset$, $A \cap C = \emptyset$. (2) Exhaustive means together they cover $S$: $A \cup B \cup C = S$.

Solution: For events on rolling a die with $S = \{1, 2, 3, 4, 5, 6\}$: Mutually exclusive: $A \cap B = \emptyset$ ✓ (no outcome is both even and odd). Exhaustive: $A \cup B = \{1, 2, 3, 4, 5, 6\} = S$ ✓ (every outcome is either even or odd).

Solution: When events are exhaustive ($A \cup B \cup C = S$) and mutually exclusive (pairwise disjoint), they form a partition of $S$. Since every outcome belongs to exactly one event, $P(A) + P(B) + P(C) = 1$. (Note: This is a consequence of the probability axiom that $P(S) = 1$.)

Solution: If $A \subset B$ (A is a proper subset of B or equal to B), then every outcome in $A$ is also in $B$. Therefore, the probability of $A$ occurring cannot exceed the probability of $B$ occurring. Hence, $P(A) \leq P(B)$.

Solution: The difference $A - B$ consists of all outcomes that are in $A$ but not in $B$. This is equivalent to $A \cap B^c$ (outcomes in $A$ AND not in $B$).

Solution: Consider $A \cap (B - A) = A \cap (B \cap A^c) = (A \cap B) \cap A^c = (A \cap A^c) \cap B = \emptyset \cap B = \emptyset$. Thus, $A$ and $B - A$ have no outcomes in common and are mutually exclusive.

Solution: We can partition $A \cup B$ into two disjoint parts: (1) $A$ (all outcomes in $A$), and (2) $B - A$ (outcomes in $B$ but not in $A$). Together, $A \cup (B - A) = A \cup B$.

Solution: Given: $S$ has 5 outcomes, $|A| = 2$, $|B| = 3$, and $|A \cap B| = 1$. Since $A \cap B \neq \emptyset$, they are NOT mutually exclusive. Also, $A \cup B$ has $2 + 3 - 1 = 4$ outcomes, which is not all 5, so they are NOT exhaustive.

Solution: There are 4 kings in a deck. Drawing 2 cards, the number of ways to get both kings (when order doesn't matter) is $\binom{4}{2} = 6$. Note: If order matters, it would be $4 \times 3 = 12$.

Solution: Compute: $A - B = \{1, 3, 5, 7\} - \{2, 4, 6, 8\} = \{1, 3, 5, 7\} = A$ (since $A$ and $B$ are disjoint). Similarly, $B - A = B$. Therefore, $(A - B) \cup (B - A) = A \cup B = \{1, 2, 3, 4, 5, 6, 7, 8\} = S$.

Solution: If $A$, $B$, $C$ partition $S$, then they are pairwise disjoint: $A \cap B = \emptyset$ and $A \cap C = \emptyset$. Therefore, $A \cap (B \cup C) = (A \cap B) \cup (A \cap C) = \emptyset \cup \emptyset = \emptyset$.

Solution: An event is any subset of $S$. The number of 3-element subsets of a 5-element set is $\binom{5}{3} = \frac{5!}{3! \cdot 2!} = 10$.

Solution: The symmetric difference contains elements in $A$ or $B$ but not both. This is $(A - B) \cup (B - A) = (A \cap B^c) \cup (A^c \cap B) = (A \cup B) - (A \cap B) = (A \cup B) \cap (A \cap B)^c$.

Solution: If $A \subseteq B$, then: (1) $A \cap B = A$ ✓, (2) $A \cup B = B$ ✓, (3) $B^c \subseteq A^c$ ✓ (complement property). However, $B - A = B \cap A^c$ need not equal $A^c$ unless $B = S$. For example, if $A = \{1\}$, $B = \{1, 2\}$, $S = \{1, 2, 3\}$, then $B - A = \{2\} \neq \{2, 3\} = A^c$.