Probability
Independent events and conditional probability
Grade 12
Question:
<p>Let \(A\) and \(B\) are two independent events such that \(P(A) = 1/3\) and \(P(B) = 1/4\). Then match the following lists:</p><table><tr><th>List I</th><th>List II</th></tr><tr><td>a. \(P(A \cup B)\) is equal to</td><td>p. 1/12</td></tr><tr><td>b. \(P(A|A \cup B)\) is equal to</td><td>q. 1/2</td></tr><tr><td>c. \(P(B/A' \cap B')\) is equal to</td><td>r. 2/3</td></tr><tr><td>d. \(P(A'/B)\) is equal to</td><td>s. 0</td></tr></table><p><strong>Codes</strong></p><table><tr><th></th><th>a</th><th>b</th><th>c</th><th>d</th></tr><tr><td>(1)</td><td>q</td><td>s</td><td>s</td><td>r</td></tr><tr><td>(2)</td><td>q</td><td>r</td><td>s</td><td>r</td></tr><tr><td>(3)</td><td>q</td><td>s</td><td>r</td><td>p</td></tr><tr><td>(4)</td><td>r</td><td>s</td><td>p</td><td>q</td></tr></table>
<p>(1) a-q, b-s, c-s, d-r</p>
<p>(2) a-q, b-r, c-s, d-r</p>
<p>(3) a-q, b-s, c-r, d-p</p>
<p>(4) a-r, b-s, c-p, d-q</p>
Step-by-Step Solution
Key Concept: For independent events, use P(A∩B) = P(A)·P(B), and carefully apply conditional probability formulas: P(X|Y) = P(X∩Y)/P(Y). Recognize when conditional events have zero probability (e.g., P(B|A'∩B') is undefined since A'∩B' and B are mutually exclusive).
Given $P(A) = 1/3$ and $P(B) = 1/4$. Events $A$ and $B$ are independent.
**Part (a): $P(A \cup B)$**
Since $A$ and $B$ are independent, $P(A \cap B) = P(A)P(B)$.
$$P(A \cup B) = P(A) + P(B) - P(A \cap B)$$
$$P(A \cup B) = P(A) + P(B) - P(A)P(B)$$
$$P(A \cup B) = \frac{1}{3} + \frac{1}{4} - \left(\frac{1}{3}\right)\left(\frac{1}{4}\right)$$
$$P(A \cup B) = \frac{1}{3} + \frac{1}{4} - \frac{1}{12}$$
$$P(A \cup B) = \frac{4}{12} + \frac{3}{12} - \frac{1}{12}$$
$$P(A \cup B) = \frac{6}{12} = \frac{1}{2}$$
**Part (b): $P(A|A \cup B)$**
By the definition of conditional probability:
$$P(A|A \cup B) = \frac{P(A \cap (A \cup B))}{P(A \cup B)}$$
Since $A \cap (A \cup B) = A$, the expression simplifies to:
$$P(A|A \cup B) = \frac{P(A)}{P(A \cup B)}$$
Using the value of $P(A \cup B)$ from Part (a):
$$P(A|A \cup B) = \frac{1/3}{1/2} = \frac{1}{3} \cdot 2 = \frac{2}{3}$$
**Part (c): $P(B|A' \cap B')$**
By the definition of conditional probability:
$$P(B|A' \cap B') = \frac{P(B \cap (A' \cap B'))}{P(A' \cap B')}$$
The event $B \cap (A' \cap B')$ implies that event $B$ occurs and event $B'$ occurs simultaneously. This is an impossible event, so $B \cap (A' \cap B') = \emptyset$.
Therefore, $P(B \cap (A' \cap B')) = P(\emptyset) = 0$.
Thus,
$$P(B|A' \cap B') = \frac{0}{P(A' \cap B')} = 0$$
(Note that $P(A' \cap B') = P((A \cup B)') = 1 - P(A \cup B) = 1 - 1/2 = 1/2 \neq 0$).
**Part (d): $P(A'|B)$**
Since $A$ and $B$ are independent events, it follows that $A'$ and $B$ are also independent events.
Therefore, by the definition of independence, $P(A'|B) = P(A')$.
$$P(A') = 1 - P(A)$$
$$P(A') = 1 - \frac{1}{3} = \frac{2}{3}$$
Correct Answer: 1