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Properties of Set Operation

An urn contai...

Question

An urn contains fair tickets with numbers 112, 121, 211, 222 and one ticket is drawn. Let $A_{i}$ (i = 1,2,3) be the event that the i $^{t h}$ digit of the number of ticket drawn is 1 then

$P (A_{1}) = P (A_{2}) = P (A_{3}) = \frac{1}{2}$

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$P (A_{1} \cap A_{2}) = P (A_{1}) P (A_{2})$

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$A_{1}, A_{2}, A_{3}$ are pair wise independent events

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$P (A_{1} \cap A_{2} \cap A_{3}) \neq P (A_{1}) P (A_{2}) P (A_{3})$

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Solution

The correct options are
A
$P (A_{1}) = P (A_{2}) = P (A_{3}) = \frac{1}{2}$

B
$P (A_{1} \cap A_{2}) = P (A_{1}) P (A_{2})$

C
$A_{1}, A_{2}, A_{3}$ are pair wise independent events

$P (A_{1} \cap A_{2}) = \frac{1}{4}, P (A_{1}) = \frac{2}{4}$

$P (A_{2}) = \frac{1}{2}$ . Thus $P (A_{1} \cap A_{2}) = P (A_{1}) P (A_{2})$

Suggest Corrections

Similar questions

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

A_{1}, A_{2}, A_{3} .

S

S = a_{1}, a_{2}, a_{3}, a_{4}

P (a_{1}) = 0.3. P (a_{2}) = 0.1

P (a_{3}) = 0.4

P (a_{4}) = 0.2

A

B

A = a_{1}, a_{2}

B = a_{2}, a_{3} a_{4}

P (A^{'} ⋂ B^{'}) =

112, 121, 211, 222

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

i^{t h}

2

p (a_{1} + \frac{1}{p q a_{2} +} \frac{1}{a_{3} +} \frac{1}{p q a_{4} +} . . . .) = p a_{1} + \frac{1}{q a_{2} +} \frac{1}{p a_{3} +} \frac{1}{q a_{4} +} . . . . .

112, 121, 211, 222

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

i^{t h}

2,

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