Question:
A box contains 6 cards numbered 1 to 6. A student is asked to pick up two cards, one by one after replacement and note down the numbers on the cards. Let A be the event of getting sum of the numbers on two cards as 10, and B, the event of a number other than 4 on the first card selected.
Find P(A and B) and find whether the events A and B are independent events or not.
A box contains 6 cards numbered 1 to 6. A student is asked to pick up two cards, one by one after replacement and note down the numbers on the cards. Let A be the event of getting sum of the numbers on two cards as 10, and B, the event of a number other than 4 on the first card selected.
Find P(A and B) and find whether the events A and B are independent events or not.
Find P(A and B) and find whether the events A and B are independent events or not.
Show Hint
In 'with replacement' problems, the denominator remains constant for each draw.
Always verify independence numerically; do not rely on intuition as events can sometimes be independent in non-obvious ways.
Always verify independence numerically; do not rely on intuition as events can sometimes be independent in non-obvious ways.
Updated On: Sep 10, 2026
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Solution and Explanation
Concept:
• Sample space with replacement: Since 2 cards are picked from 6 with replacement, \( n(S) = 6 \times 6 = 36 \).
• Probability of intersection: \( P(A \cap B) = \frac{n(A \cap B)}{n(S)} \).
• Independent Events: A and B are independent if and only if \( P(A \cap B) = P(A) \cdot P(B) \).
Step 1: List the outcomes for event A (Sum = 10)
Possible pairs \( (x, y) \) where \( x+y = 10 \) and \( x, y \in \{1, 2, 3, 4, 5, 6\} \): \( A = \{(4, 6), (5, 5), (6, 4)\} \). \( n(A) = 3 \). \[ P(A) = \frac{3}{36} = \frac{1}{12} \]
Step 2: Analyze event B (First card \( \neq 4 \)) and find \( P(B) \)
The first card can be any of \( \{1, 2, 3, 5, 6\} \) (5 choices). The second card can be any of \( \{1, 2, 3, 4, 5, 6\} \) (6 choices). \( n(B) = 5 \times 6 = 30 \). \[ P(B) = \frac{30}{36} = \frac{5}{6} \]
Step 3: Find \( P(A \cap B) \)
\( A \cap B \) means the sum is 10 AND the first card is not 4. From the set \( A \): - \( (4, 6) \): First card is 4 (Discard) - \( (5, 5) \): First card is not 4 (Keep) - \( (6, 4) \): First card is not 4 (Keep) So, \( A \cap B = \{(5, 5), (6, 4)\} \Rightarrow n(A \cap B) = 2 \). \[ P(A \cap B) = \frac{2}{36} = \frac{1}{18} \]
Step 4: Check for independence
Check if \( P(A \cap B) = P(A) \cdot P(B) \): \[ P(A) \cdot P(B) = \frac{1}{12} \times \frac{5}{6} = \frac{5}{72} \] Since \( \frac{1}{18} = \frac{4}{72} \) and \( \frac{4}{72} \neq \frac{5}{72} \), we have: \[ P(A \cap B) \neq P(A) \cdot P(B) \] Thus, events A and B are not independent (they are dependent).
• Sample space with replacement: Since 2 cards are picked from 6 with replacement, \( n(S) = 6 \times 6 = 36 \).
• Probability of intersection: \( P(A \cap B) = \frac{n(A \cap B)}{n(S)} \).
• Independent Events: A and B are independent if and only if \( P(A \cap B) = P(A) \cdot P(B) \).
Step 1: List the outcomes for event A (Sum = 10)
Possible pairs \( (x, y) \) where \( x+y = 10 \) and \( x, y \in \{1, 2, 3, 4, 5, 6\} \): \( A = \{(4, 6), (5, 5), (6, 4)\} \). \( n(A) = 3 \). \[ P(A) = \frac{3}{36} = \frac{1}{12} \]
Step 2: Analyze event B (First card \( \neq 4 \)) and find \( P(B) \)
The first card can be any of \( \{1, 2, 3, 5, 6\} \) (5 choices). The second card can be any of \( \{1, 2, 3, 4, 5, 6\} \) (6 choices). \( n(B) = 5 \times 6 = 30 \). \[ P(B) = \frac{30}{36} = \frac{5}{6} \]
Step 3: Find \( P(A \cap B) \)
\( A \cap B \) means the sum is 10 AND the first card is not 4. From the set \( A \): - \( (4, 6) \): First card is 4 (Discard) - \( (5, 5) \): First card is not 4 (Keep) - \( (6, 4) \): First card is not 4 (Keep) So, \( A \cap B = \{(5, 5), (6, 4)\} \Rightarrow n(A \cap B) = 2 \). \[ P(A \cap B) = \frac{2}{36} = \frac{1}{18} \]
Step 4: Check for independence
Check if \( P(A \cap B) = P(A) \cdot P(B) \): \[ P(A) \cdot P(B) = \frac{1}{12} \times \frac{5}{6} = \frac{5}{72} \] Since \( \frac{1}{18} = \frac{4}{72} \) and \( \frac{4}{72} \neq \frac{5}{72} \), we have: \[ P(A \cap B) \neq P(A) \cdot P(B) \] Thus, events A and B are not independent (they are dependent).
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