A coin is tossed four times. Consider the following events.
$E_{1} :$ Head on the first toss.
$E_{2} :$ Tails on the second toss.
$E_{3} :$ Heads on third toss.
$E_{4} :$ All outcomes are same
$E_{5} :$ Exactly one head turns up.
$E_{6} :$ Tails on fourth toss.
$E_{7} :$ Exactly on tails turns up
Which of the following sets of events is/are independent?

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$E_{1} a n d E_{4}$

$E_{1} a n d E_{5}$

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

$E_{1}, E_{2}, E_{6} a n d E_{7}$

answer is A, C.

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Detailed Solution

$E_{1} = {H H T, H T H, H T T, H H H} \to \frac{1}{2} E_{2} = {H T T, T T H, H T H, T T T} \to \frac{1}{2} E_{3} = {H T H, T H H, H H H, T T H} \to \frac{1}{2} E_{4} = {H H H, T T T} \to \frac{1}{4} E_{5} = {H T T, T H T, T T H} \to \frac{3}{8} P (E_{1} \cap E_{4}) = P {H H H} = \frac{1}{8} = P (E_{1}) . P (E_{4}) \Rightarrow (A) P (E_{1} \cap E_{5}) = P {H T T} = \frac{1}{8} \neq P (E_{1}) . P (E_{5}) P (E_{1} \cap E_{2}) = P {H H T, H T H} = \frac{1}{4} = P (E_{1}) . P (E_{2}) P (E_{2} \cap E_{3}) = P {T T H, H T H} = \frac{1}{4} = P (E_{2}) . P (E_{3}) P (E_{3} \cap E_{1}) = P {H T H, H H H} = \frac{1}{4} = P (E_{3}) . P (E_{1}) P (E_{1} \cap E_{2} \cap E_{3}) = P {H T H} = \frac{1}{8} = P (E_{1}) . P (E_{2}) . P (E_{3}) \Rightarrow E_{1}, E_{2}, E_{3} an independent \Rightarrow (C)$