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Step-by-Step Solution
Step 1: Identify the total number of outcomes
When two fair six-faced dice (say, A and B) are thrown, each die has 6 faces, so
the total number of possible outcomes is
$6 \times 6 = 36$.
Step 2: Define the events
$E_1$: Die A shows 4.
Possible outcomes for $E_1$:
{(4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6)}
Number of favorable outcomes = 6
$P(E_1) = \frac{6}{36} = \frac{1}{6}$
$E_2$: Die B shows 2.
Possible outcomes for $E_2$:
{(1, 2), (2, 2), (3, 2), (4, 2), (5, 2), (6, 2)}
Number of favorable outcomes = 6
$P(E_2) = \frac{6}{36} = \frac{1}{6}$
$E_3$: The sum of numbers on both dice is odd.
An odd sum occurs when one die shows an even number and the other shows an odd number.
The number of such outcomes is 18 (half of the total 36).
$P(E_3) = \frac{18}{36} = \frac{1}{2}$
Step 3: Check pairwise independence
(a) Independence of $E_1$ and $E_2$
$E_1 \cap E_2$ means die A shows 4 and die B shows 2. There is exactly 1 outcome: (4, 2).
$P(E_1 \cap E_2) = \frac{1}{36}$
$P(E_1)\cdot P(E_2) = \frac{1}{6} \times \frac{1}{6} = \frac{1}{36}$
Since $P(E_1 \cap E_2) = P(E_1) \times P(E_2)$, $E_1$ and $E_2$ are independent.
(b) Independence of $E_1$ and $E_3$
$E_1 \cap E_3$ means die A shows 4 and the sum of both dice is odd. The outcomes satisfying this are {(4, 1), (4, 3), (4, 5)}.
Number of favorable outcomes = 3
$P(E_1 \cap E_3) = \frac{3}{36} = \frac{1}{12}$
$P(E_1)\cdot P(E_3) = \frac{1}{6} \times \frac{1}{2} = \frac{1}{12}$
Since $P(E_1 \cap E_3) = P(E_1) \times P(E_3)$, $E_1$ and $E_3$ are independent.
(c) Independence of $E_2$ and $E_3$
$E_2 \cap E_3$ means die B shows 2 and the sum of both dice is odd. The outcomes satisfying this are {(1, 2), (3, 2), (5, 2)}.
Number of favorable outcomes = 3
$P(E_2 \cap E_3) = \frac{3}{36} = \frac{1}{12}$
$P(E_2)\cdot P(E_3) = \frac{1}{6} \times \frac{1}{2} = \frac{1}{12}$
Since $P(E_2 \cap E_3) = P(E_2)\times P(E_3)$, $E_2$ and $E_3$ are independent.
Step 4: Check the three-way independence
For three events $E_1, E_2, E_3$ to be mutually independent, we also need:
$P(E_1 \cap E_2 \cap E_3) = P(E_1)\times P(E_2)\times P(E_3)$.
The event $E_1 \cap E_2 \cap E_3$ means:
"Die A shows 4, Die B shows 2, and the sum is odd."
But (4 + 2) = 6, which is even. Hence there is no outcome where A is 4, B is 2, and the total is odd.
Therefore the intersection is empty, and:
$P(E_1 \cap E_2 \cap E_3) = 0.$
On the other hand,
$P(E_1)\,P(E_2)\,P(E_3) = \frac{1}{6} \times \frac{1}{6} \times \frac{1}{2} = \frac{1}{72} \neq 0.$
Since these two are not equal, $E_1, E_2,$ and $E_3$ are not mutually independent.
Step 5: Identify the statement that is NOT true
From the above steps, we see that $E_1, E_2,$ and $E_3$ are not mutually independent. Therefore, the statement
“$E_1, E_2,$ and $E_3$ are independent” is not true.
Final Answer
The statement (Option 4) $E_1, E_2,$ and $E_3$ are independent is not true.
