Question

A fair coin is tossed three times. Consider the following events.
$E_1:$ Heads on the first toss.
$E_2:$ Tails on the second toss.
$E_3:$ Heads on the third toss.
$E_4:$ All three outcomes are same.
$E_5:$ Exactly one heads turns up.

Which of the following pairs/triples of events is (are) independent?

• A. $E_1$ and $E_4$
• B. $E_1$ and $E_5$
• C. $E_1,E_2$ and $E_3$
• D. $E_1,E_2$ and $E_4$

Answer 1

Answer: (A), (C)

$E_1,E_2$ and $E_3$

$E_1=\{HHT,HTH,HTT,HHH\}$
$P\left(E_1\right)=\frac{4}{8}=\frac{1}{2}$
$E_2=\{HTT,TTH,HTH,TTT\}$
$P\left(E_2\right)=\frac{4}{8}=\frac{1}{2}$
$E_3=\{HTH,THH,HHH,TTH\}$
$P\left(E_3\right)=\frac{4}{8}=\frac{1}{2}$
$E_4=\{HHH,TTT\}$
$P\left(E_4\right)=\frac{2}{8}=\frac{1}{4}$
$E_5=\{HTT,THT,TTH\}$
$P\left(E_5\right)=\frac{3}{8}$

$P(E_1\cap E_4)=P\left\{HHH\right\}=\frac{1}{8}=P\left(E_1\right)\cdot P\left(E_4\right),$ hence independent.

$P(E_1\cap E_5)=P\left\{HTT\right\}=\frac{1}{8}\ne P\left(E_1\right)\cdot P\left(E_5\right),$ hence not independent.

$P(E_1\cap E_2\cap E_3)=P\left\{HTH\right\}=\frac{1}{8}=P\left(E_1\right)\cdot P\left(E_2\right)\cdot P\left(E_3\right)$
$\Rightarrow E_1,E_2,E_3$ are independent.

Obviously, $E_1\cap E_2\cap E_4=\varphi ,$ hence not independent.