Q.For a loaded die, the probabilities of outcomes are given as under: $P(1) = P(2) = 0.2$, $P(3) = P(5) = P(6) = 0.1$ and $P(4) = 0.3$. The die is thrown two times. Let $A$ and $B$ be the events, 'same number each time', and 'a total score is $10$ or more', respectively. Determine whether or not $A$ and $B$ are independent.

Concept: Event Independence — Two events $A$ and $B$ are independent iff $P(A \cap B) = P(A) \cdot P(B)$.

Step 1: Find $P(A)$

$A$ = same number both times.

$P(A) = \sum_{i=1}^6 [P(i)]^2 = (0.2)^2 + (0.2)^2 + (0.1)^2 + (0.3)^2 + (0.1)^2 + (0.1)^2$

$= 0.04 + 0.04 + 0.01 + 0.09 + 0.01 + 0.01 = 0.20$.

Step 2: Find $P(B)$

$B$ = total score $\ge 10$. Pairs: (4,6), (5,5), (5,6), (6,4), (6,5), (6,6).

$P(4,6) = 0.3 \times 0.1 = 0.03$, $P(5,5) = 0.1 \times 0.1 = 0.01$, $P(5,6) = 0.1 \times 0.1 = 0.01$,

$P(6,4) = 0.1 \times 0.3 = 0.03$, $P(6,5) = 0.1 \times 0.1 = 0.01$, $P(6,6) = 0.1 \times 0.1 = 0.01$.

Sum = $0.03+0.01+0.01+0.03+0.01+0.01 = 0.10$.

Step 3: Find $P(A \cap B)$

$A \cap B$ = same number both times and total $\ge 10$. Only (5,5) and (6,6) qualify.

$P(5,5) = 0.01$, $P(6,6) = 0.01$, so $P(A \cap B) = 0.02$.

Step 4: Check independence

$P(A) \cdot P(B) = 0.20 \times 0.10 = 0.02$.

Since $P(A \cap B) = 0.02 = P(A) \cdot P(B)$, the events are independent.

Two events are independent exactly when $P(A\cap B)=P(A)\,P(B)$. The two throws are independent, so the probability of any ordered pair $(x,y)$ is $P(x)\,P(y)$.

Given single-throw probabilities.

(They sum to $1$.)

1. $P(A)$ — same number both times. This is $\sum_k P(k)^2$:

2. $P(B)$ — total score $\ge 10$. The ordered pairs and their probabilities:

• total $10$: $(4,6)=0.03$, $(6,4)=0.03$, $(5,5)=0.01$
• total $11$: $(5,6)=0.01$, $(6,5)=0.01$
• total $12$: $(6,6)=0.01$

3. $P(A\cap B)$ — same number and total $\ge 10$. Equal faces give total $2k$, which is $\ge 10$ only for $k=5$ or $k=6$:

4. Test independence.

The two sides agree, so the condition for independence is met.

Method: Testing Whether Two Events Are Independent

Use this whenever a problem asks "are $A$ and $B$ independent?" — the answer is always decided by one equation, never by intuition.

Steps

Step 1: Compute the three probabilities $P(A)$, $P(B)$, $P(A\cap B)$ from the experiment.

For two throws of a die the throws are independent, so any ordered pair $(x,y)$ has probability $P(x)\,P(y)$. Build each event by listing the qualifying pairs and adding their probabilities. (For a "same number both times" event, $P(A)=\sum_k P(k)^2$.)

Step 2: Apply the independence test.

Step 3: Compare and conclude.

Compute $P(A)\,P(B)$ and compare with the directly-counted $P(A\cap B)$. Equal $\Rightarrow$ independent; unequal $\Rightarrow$ not independent. There is no "close enough" — it must be exact.

Common Mistakes

Mistake 1: Taking $P(A)=\tfrac16$ ("same number") as if the die were fair.

Why it's wrong: the die is loaded, so "same number both times" is $\sum_k P(k)^2=0.20$, not $\tfrac16$. Correct approach: square each face probability and add.

Mistake 2: Adding face probabilities for a pair instead of multiplying.

Why it's wrong: the two throws are independent, so $P(x,y)=P(x)P(y)$; e.g. $P(4,6)=0.3\times0.1=0.03$. Correct approach: multiply per-throw probabilities.

Mistake 3: Concluding dependence from intuition without the test.

Why it's wrong: independence is decided only by $P(A\cap B)=P(A)P(B)$; here $0.02=0.20\times0.10$, so they are independent even though it may feel otherwise.