A bag contains four tickets marked with $112, 121, 211, 222$ , one ticket is drawn at random from the bag. Let $E_{i} (i = 1, 2, 3)$ denote the event that $i^{t h}$ digit on the ticket is $2$ then :

E_{1}

and

E_{2}

are independent

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E_{2}

and

E_{3}

are independent

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E_{3}

and

E_{1}

are independent

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E_{1}, E_{2}, E_{2}

are independent

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Solution

The correct options are
A $E_{1}$ and $E_{2}$ are independent
B $E_{2}$ and $E_{3}$ are independent
C $E_{3}$ and $E_{1}$ are independent
$P (E_{1}) = P (E_{2}) = P (E_{3}) = \frac{1}{2}$

$P (E_{i} \cap E_{j}) = \frac{1}{4} = P (E_{i}) P (E_{j})$

Hence, two events taken together are independent.

$P (E_{1} \cap E_{2} \cap E_{3}) = \frac{1}{4} \neq P (E_{1}) P (E_{2}) P (E_{3})$

Therefore, the three events are not independent together.

Hence, options A, B and C are correct.

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Similar questions

Q. A bag contains four tickets marked 112, 121, 211, 222. One ticket is drawn at random from the bag.
Let

E_{i} (i = 1, 2, 3)

denote the event that

i^{t h}

digit on the ticket is 2. Which of the following option(s) is/are correct?

Q. If

P (E_{1}) = 0.2, P (E_{2}) = 0.4

and

P (E_{3}) = 0.6

and

E_{1}, E_{2}

and

E_{3}

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E_{1}, E_{2}

and

E_{3}

occurs is

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