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 \( P(A \text{ and } B) \) and find whether the events A and B are independent events or not.

Hint:

With replacement means the same outcome can repeat (like 5,5). Always list the intersection by looking at the smaller set (usually the one with more restrictions) and checking if those elements fit the other set's rule.

Answer 1

Concept:
• Sample Space \( S \): Total number of possible outcomes.
• Event Intersection \( A \cap B \): Outcomes that satisfy both conditions simultaneously.
• Condition for Independence: Two events \( A \) and \( B \) are independent if and only if \( P(A \cap B) = P(A) \cdot P(B) \).

Step 1: Determine the total number of outcomes in the sample space
There are 6 cards numbered \( \{1, 2, 3, 4, 5, 6\} \).
Two cards are picked "one by one after replacement".
Total outcomes \( n(S) = 6 \times 6 = 36 \).

Step 2: Define event A and calculate its probability
Event A is the event of getting a sum of 10.
Possible pairs \( (x, y) \) such that \( x + y = 10 \) where \( x, y \in \{1, 2, 3, 4, 5, 6\} \):
\[ A = \{(4, 6), (5, 5), (6, 4)\} \]
Number of outcomes in A, \( n(A) = 3 \).
Probability \( P(A) = \frac{n(A)}{n(S)} = \frac{3}{36} = \frac{1}{12} \).

Step 3: Define event B and calculate its probability
Event B is the event that a number other than 4 appears on the first card.
The first card can be any number from \( \{1, 2, 3, 5, 6\} \) (5 choices).
The second card can be any number from \( \{1, 2, 3, 4, 5, 6\} \) (6 choices).
Number of outcomes in B, \( n(B) = 5 \times 6 = 30 \).
Probability \( P(B) = \frac{n(B)}{n(S)} = \frac{30}{36} = \frac{5}{6} \).

Step 4: Find \( P(A \cap B) \)
\( A \cap B \) contains outcomes from A where the first card is not 4.
From set A: \( (4, 6) \) starts with 4 (excluded), \( (5, 5) \) and \( (6, 4) \) do not start with 4.
\[ A \cap B = \{(5, 5), (6, 4)\} \]
Number of outcomes in \( A \cap B \), \( n(A \cap B) = 2 \).
Probability \( P(A \text{ and } B) = \frac{2}{36} = \frac{1}{18} \).

Step 5: Check for independence
Calculate the product \( P(A) \cdot P(B) \):
\[ P(A) \cdot P(B) = \frac{1}{12} \cdot \frac{5}{6} = \frac{5}{72} \]
Compare with \( P(A \cap B) \):
\[ P(A \cap B) = \frac{1}{18} = \frac{4}{72} \]
Since \( \frac{4}{72} \neq \frac{5}{72} \), we have \( P(A \cap B) \neq P(A) \cdot P(B) \).
Therefore, events A and B are not independent.