Let two fair six-faced dice A and B be thrown simulatneously. If $E_{1}$ is the event that die A shows up four, $E_{2}$ is the event that die B shows up two and $E_{3}$ is the event that the sum of numbers on both dice is odd, then which of the following statements is/are true ?

$E_{1} a n d E_{2}$ are independent

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$E_{2} a n d E_{3}$ are independent

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$E_{1} a n d E_{3}$ are independent

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$E_{1}, E_{2} a n d E_{3}$ are independent

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Solution

The correct options are
A
$E_{1} a n d E_{2}$ are independent

B
$E_{2} a n d E_{3}$ are independent

C
$E_{1} a n d E_{3}$ are independent

Clearly, $E_{1} = {(4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6)} E_{2} = {(1, 2), (2, 2), (3, 2), (4, 2), (5, 2), (6, 2)} a n d E_{3} = {(1, 2), (1, 4), (1, 6), (2, 1), (2, 3), (2, 5) (3, 2), (3, 4), (3, 6), (4, 1), (4, 3), (4, 5), (5, 2), (5, 4), (5, 6), (6, 1), (6, 3), (6, 5)}$
$\Rightarrow P (E_{1}) = \frac{6}{36} = \frac{1}{6}, P (E_{2}) = \frac{6}{36} = \frac{1}{6}$
$a n d P (E_{3}) = \frac{18}{36} = \frac{1}{2}$
Now, $P (E_{1} \cap E_{2}) = P$
(getting 4 on die A and 2 on die B)
$= \frac{1}{36} = \frac{1}{6} \times \frac{1}{6} = P (E_{1}) . P (E_{2})$
$P (E_{2} \cap E_{3}) = P$ (getting 2 on die B and sum of numbers on both dice is odd)
$= \frac{3}{36} = \frac{1}{12} = \frac{1}{6} \times \frac{1}{2} = P (E_{2}) . P (E_{3})$
$P (E_{1} \cap E_{3}) = P$ (getting 4 on die A and sum of numbers on both dice is odd)
$= \frac{3}{36} = \frac{1}{12} = \frac{1}{6} \times \frac{1}{2} = P (E_{1}) . P (E_{3})$
and $P (E_{1} \cap E_{2} \cap E_{3}) = P$ (getting 4 on die A, 2 on die B and sum of numbers is odd)
=P (impossible event )=0 $\neq P (E_{1}) . P (E_{2}) . P (E_{3})$
Hence, $E_{1}, E_{2} a n d E_{3}$ are not independent.

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Let two fair six-faced dice A and B be thrown simulatneously. If $E_{1}$ is the event that die A shows up four, $E_{2}$ is the event that die B shows up two and $E_{3}$ is the event that the sum of numbers on both dice is odd, then which of the following statements is/are true ?

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