7.0 Independent Events

7.1 Occurrence of at least one of the independent events
7.2 Pairwise independent events
7.3 Mutually independent events

Events are said to be independent when the happening of any one event does not affect the happening of any of the others.

i. $A$ and $B$ are independent events if

$$P(A/B) = P(A)$$

or

$$P(B/A) = P(B)$$

Proof: If $A$ and $B$ are independent events, then $P(A/B) = {{P(A \cap B)} \over {P(B)}} = {{P(A)P(B)} \over {P(B)}} = P(A)$ ,this is true as occurrence of $A$ is independent of event $B$.

ii. The probability of the occurrence of several independent events is the product of probability of individual events.

i.e. $$P({E_1} \cap {E_2} \cap .... \cap {E_n}) = P({E_1})P({E_2})P({E_3})......P({E_n})$$

Illustration 28. A coin is tossed thrice and all outcomes are equally likely. If the events are, getting head in the first throw and getting tail in the last throw, show that they are independent.

Solution:

Consider the events,

$A$ - Getting head in the first throw.

$B$ - Getting tail in the last throw.

The sample space for tossing a coin thrice is

$$S = \{ (HHH),(HHT),(HTH),(THH),(HTT),(TTH),(THT),(TTT)\} $$

$A = \{ (HHH),(HHT),(HTH),(HTT)\} $

$B = \{ (HHT),(HTT),(THT),(TTT)\} $

$A \cap B = \{ (HHT),(HTT)\} $

$$\begin{equation} \begin{aligned} n(A) = 4 \\ n(B) = 4 \\ n(A \cap B) = 2 \\ P(A) = {{n(A)} \over {n(S)}} = {4 \over 8} = {1 \over 2} \\ P(B) = {{n(A)} \over {n(S)}} = {4 \over 8} = {1 \over 2} \\\end{aligned} \end{equation} $$

$$P(A \cap B) = {{n(A \cap B)} \over {n(S)}} = {2 \over 8} = {1 \over 4} = {1 \over 2} \times {1 \over 2} = P(A) \times P(B)$$

Hence, $A$ and $B$ are independent events.

Illustration 29. Given that the events $A$ and $B$ are such that $P(A) = {1 \over 2}$, $P(A \cup B) = {3 \over 5}$ and $P(B) = p$ . Find $p$ if they are independent.

Solution: If $A$ and $B$ are independent events,

$$\begin{equation} \begin{aligned} P(A \cup B) = P(A) + P(B) - P(A \cap B) \\ P(A \cap B) = P(A) \times P(B) \\ P(A \cap B) = {1 \over 2}p \\ {3 \over 5} = {1 \over 2} + p - {1 \over 2}p \\ {1 \over 2}p = {3 \over 5} - {1 \over 2} \\ p = {1 \over 5} \\\end{aligned} \end{equation} $$

Illustration 30. If $A$ and $B$ are two events such that $P(A) = {1 \over 4}$, $P(A \cup B) = {1 \over 8}$ and $P(B) = {1 \over 2}$, find $P(not\;A\;and\;not\;B)$.

Solution: $P(A \cap B) = {1 \over 8} = {1 \over 4} \times {1 \over 2} = P(A) \times P(B)$

So, $A$ and $B$ are independent events.

Now,

$$\begin{equation} \begin{aligned} P(not\;A\;and\;not\;B) \\ = P(\bar A \cap \bar B) \\ = P(\bar A) \times P(\bar B) \\ = \left( {1 - P(A)} \right) \times \left( {1 - P(B)} \right) \\ = \left( {1 - {1 \over 4}} \right) \times \left( {1 - {1 \over 2}} \right) \\ = {3 \over 4} \times {1 \over 2} \\ = {3 \over 8} \\\end{aligned} \end{equation} $$