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 the sum of the numbers on the two cards as 10, and $B$ be 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:

When computing sample spaces with replacement, the denominator is always $n^k$. Always look out for words like "with replacement" or "without replacement" as they entirely alter the conditional probabilities of subsequent events!

Answer 1

Concept: Two events $A$ and $B$ associated with a random experiment are defined to be mathematically independent if and only if the probability of their simultaneous occurrence satisfies the product rule: \[ P(A \cap B) = P(A) \cdot P(B) \] If this multiplicative equation does not hold true, the events are classified as dependent. Here, two cards are drawn sequentially with replacement from a set of 6 unique cards, meaning each draw has 6 independent outcomes.

Step 1: Determining the total sample space and listing outcomes for Event $A$.
Since two cards are drawn one by one with replacement from a box containing cards numbered $\{1, 2, 3, 4, 5, 6\}$, the total number of elementary outcomes in the sample space $S$ is: \[ n(S) = 6 \times 6 = 36 \] Event $A$ is defined as getting a sum equal to 10 on the two drawn cards. Let us systematically list all ordered pairs $(x, y)$ such that $x + y = 10$, where $1 \leq x, y \leq 6$:
• If $x = 4$, then $y = 6 \implies (4, 6)$
• If $x = 5$, then $y = 5 \implies (5, 5)$
• If $x = 6$, then $y = 4 \implies (6, 4)$ Thus, $A = \{(4, 6), (5, 5), (6, 4)\}$, and the number of favorable outcomes is $n(A) = 3$. The probability of event $A$ is: \[ P(A) = \frac{n(A)}{n(S)} = \frac{3}{36} = \frac{1}{12} \]

Step 2: Listing outcomes and finding the probability for Event $B$.
Event $B$ is defined as getting a number other than 4 on the first card selected. This means the first card can be any number from the set $\{1, 2, 3, 5, 6\}$ (5 possibilities), and the second card can be any number from $\{1, 2, 3, 4, 5, 6\}$ (6 possibilities). The total number of favorable outcomes for event $B$ is: \[ n(B) = 5 \times 6 = 30 \] The probability of event $B$ is: \[ P(B) = \frac{n(B)}{n(S)} = \frac{30}{36} = \frac{5}{6} \]

Step 3: Finding the intersection set $(A \cap B)$ and computing $P(A \text{ and } B)$.
The intersection event $(A \cap B)$ represents the outcomes where the sum of the numbers is 10 AND the first card is not equal to 4. Let us filter the elements of set $A$ to remove any pair where the first coordinate is 4:
• $(4, 6)$ has 4 as the first card $\implies$ excluded.
• $(5, 5)$ does not have 4 as the first card $\implies$ included.
• $(6, 4)$ does not have 4 as the first card $\implies$ included. Therefore, $A \cap B = \{(5, 5), (6, 4)\}$, which gives $n(A \cap B) = 2$. The probability of their simultaneous occurrence is: \[ P(A \cap B) = \frac{n(A \cap B)}{n(S)} = \frac{2}{36} = \frac{1}{18} \]

Step 4: Testing for mathematical independence.
Let us check if the product of individual probabilities equals the joint intersection probability: \[ P(A) \cdot P(B) = \frac{1}{12} \times \frac{5}{6} = \frac{5}{72} \] Comparing this with our calculated value of $P(A \cap B)$: \[ P(A \cap B) = \frac{1}{18} = \frac{4}{72} \neq \frac{5}{72} \] Since $P(A \cap B) \neq P(A) \cdot P(B)$, the events $A$ and $B$ are verified to be Dependent.