Q.Refer to Exercise 1 above. If the die were fair, determine whether or not the events $A$ and $B$ are independent.

Independence test: $A$ and $B$ are independent only if $P(A \cap B) = P(A)\,P(B)$. Reusing Exercise 1's set-up (a die thrown twice) with the events $A$: the same number comes up both times, and $B$: the total score is $10$ or more — but now the die is fair.

• $A = \{(1,1),(2,2),(3,3),(4,4),(5,5),(6,6)\}$, so $P(A) = \frac{6}{36} = \frac{1}{6}$.
• $B$ (total $\ge 10$) $= \{(4,6),(5,5),(6,4),(5,6),(6,5),(6,6)\}$, so $P(B) = \frac{6}{36} = \frac{1}{6}$.
• $A \cap B = \{(5,5),(6,6)\}$, so $P(A \cap B) = \frac{2}{36} = \frac{1}{18}$.

Compare: $P(A)\,P(B) = \frac{1}{6}\cdot\frac{1}{6} = \frac{1}{36}$, but $P(A \cap B) = \frac{1}{18}$.

What the question is asking

This is the same experiment as Exercise 1 — a die thrown two times, with

• $A$ = the same number turns up on both throws,
• $B$ = the total of the two throws is $10$ or more—

but here we assume the die is fair, and we must decide whether $A$ and $B$ are independent. Two events are independent precisely when

So we compute all three probabilities and check the equality.

Step 1: the sample space

Throwing a fair die twice gives $6 \times 6 = 36$ equally likely ordered pairs, each with probability $\frac{1}{36}$.

Step 2: probability of $A$ (same number both times)

Step 3: probability of $B$ (total $\ge 10$)

Totals of $10$, $11$ or $12$:

Step 4: probability of $A \cap B$

We need outcomes that are in both lists — the number is the same and the total is at least $10$:

Step 5: apply the test

Since $\frac{1}{18} \ne \frac{1}{36}$, the product rule fails.

Method: Testing Independence Under a Fair (Equally-Likely) Model

Use this to decide independence when outcomes are equally likely, so every probability is just a count over the total.

Steps

Step 1: Fix the sample space.

Two throws of a fair die give $6\times6=36$ equally likely ordered outcomes, each of probability $\tfrac1{36}$.

Step 2: Count each event.

where $A\cap B$ is found by listing the outcomes lying in both events (not by multiplying).

Step 3: Apply the product-rule test.

Step 4: Conclude by exact comparison. Note that changing the die's fairness can change the verdict, so the model matters — the same events can be independent under one weighting and dependent under another.

Common Mistakes

Mistake 1: Reusing the loaded-die answer and assuming still independent.

Why it's wrong: with a fair die the numbers change — $P(A\cap B)=\frac{1}{18}$ but $P(A)P(B)=\frac{1}{36}$, so they are not independent. Correct approach: recompute every probability under the fair model.

Mistake 2: Multiplying to get $P(A\cap B)$.

Why it's wrong: that would assume the conclusion; $A\cap B$ must be found by listing shared outcomes $\{(5,5),(6,6)\}$, giving $\frac{2}{36}$. Correct approach: count the intersection directly.

Mistake 3: Calling $A$ and $B$ mutually exclusive.

Why it's wrong: they share $(5,5)$ and $(6,6)$, so they can occur together; the issue is only the product rule, not exclusivity.