In independent probability two events are said to be independent when the occurrence or non occurrence of one does not depend on the occurrence or non occurrence of the other.
Two events A and B are independent if and only if
$P(A \cap B)$ = $P(A)$ . $P(B)$
This concept can also be extended to dealing with collections of more than two events or random variables. Independent probabilities are not affected by previous probabilities. If two events are independent then they cannot be mutually exclusive (disjoint) and vice versa.

Example : John is playing cards with his friends, with replacement, and he draws a red card for the first time, then the drawing of a red card by John again is independent of the first event. Since the card drawn is again placed back in the deck of cards.

## Independent Probability Formula

When A and B are two independent events we can easily find the probability of $A$ and then $B$ happening. To find the probability of both the events we multiply the events.
Therefore, independent probability formula for event A and event B is

$P(A and B)$ = $P(A)$ $\times$ $P(B)$

## Independent Probability Rules

Certain rules need to be followed when studying probability of events.

1)  If there is more than one event we have
$P(S)$ = $P(A \cup B \cup C)$

2)  If A and B are independent events, probability of event happening is
$P(A \cap B)$ = $P(A)$ $\times$ $P(B)$

a) $P(A + B)$ = $P(A)$ + $P(B)$ $- P(A \cap B)$

b) $(A \cup B)$ = $1 - (A \cup B)'$

4)  When $A$ and $B$ are two mutually exclusive events with probabilities $P(A)$ and $P(B)$. Probability of occurrence of atleast one of these two events is

$P(A \cup B)$ = $P(A)$ + $P(B)$

5) Conditional probability for two independent events, A and B, is

$P$ $(\frac{B}{A})$ = $\frac{P(A\cap B)}{P(A)}$

## Dependent Probability

Dependent probability occurs when the first item is not replaced before drawing the second item. Here the events are said to be dependent. Occurrence of event $A$ changes the probability of event $B$, then events $A$ and $B$ are dependent.
In dependent probability one event affects the next event and what happens depends on what happened before.The formula for dependent probability of events is

$P(A and B)$ = $P(A)$ $\times$ $P$ $(\frac{B}{A})$

Example 1: A bag contains 4 red and 6 blue balls. Two balls are drawn at random from the bag. Find the probability that both of them are red if
1) The balls are drawn one after the other, without replacement.
2) The balls are drawn one after the other, with replacement.

Solution :

1)  Draw without replacement.
The balls are drawn one after the other. The balls drawn are not returned to the bag, they are kept aside.
Let $A$ : First ball drawn is red.
$B$ : Second ball drawn is red.

$P(A)$ = $\frac{4}{10}$ = $\frac{2}{5}$

$P$ $(\frac{B}{A})$ = $\frac{3}{9}$ = $\frac{1}{3}$,

because when $A$ has happened, the bag will have only 3 red and 6 blue balls.

Therefore, $P$ [Both the balls are red] = $P(A \cap B)$

= $P(A)$ . $P$ $(\frac{B}{A})$

= $\frac{2}{5}$ $\times$ $\frac{1}{3}$

= $\frac{2}{15}$

= 0.1333

2)  Draw with replacement.
The balls are drawn one after the other. However after drawing the first ball, the drawn ball is returned to the bag before the second draw is made.
Let $A$ : First ball drawn is red.
$B$ : Second ball drawn is red.

$P(A)$ = $\frac{4}{10}$ = $\frac{2}{5}$.

Contents of the bag remain unaltered as the first ball drawn is returned before the second draw.

Therefore, $P$ [Both the balls are red] = $P(A \cap B)$

= $P(A)$ . $P$ $(\frac{B}{A})$

= $\frac{2}{5}$ $\times$ $\frac{2}{5}$

= $\frac{4}{25}$

= 0.16

## Independent Probability Problems

Example1: A fair coin and a fair die are thrown. Find the probabilities of
1)  Head on the coin and the number 6 on the die.
2)  Head on the coin and even number on the die.

Solution :
Let $A$ : The coin shows head.
$B$ : The die shows the number 6.
$C$: The die shows even number.

$P(A)$ = $\frac{1}{2}$, $P(B)$ = $\frac{1}{6}$, $P(C)$ = $\frac{3}{6}$ = $\frac{1}{2}$

1) Events A and B are independent.

Therefore, $P$ [Head on the coin and 6 on the die] = $P(A \cap B)$

= $P(A)$ . $P(B)$

= $\frac{1}{2}$ . $\frac{1}{6}$

= $\frac{1}{12}$

= 0.0833

2) Events $A$ and $C$ are independent.

Therefore, $P$ [Head on the coin and even number on the die] = $P(A \cap C)$

= $P(A)$ . $P(C)$

= $\frac{1}{2}$ . $\frac{1}{2}$

= $\frac{1}{4}$

= 0.25

Example 2: A bag contains 25 chocolates, among them 17 are Dairymilk silk chocolates and 8 of the chocolates are Bournville.
1) What is the probability of getting a Bournville?
2) What is the probability of getting a Dairymilk silk?
3) What is the probability of getting Bournville and then Dairymilk silk?

Solution :
Given :
A bag contains 25 chocolates, 17 are Dairymilk  silk chocolates and 8 of them are Bournville.

1) Probability of getting a Bournville
As there are 8 Bournville chocolates among 25,

Probability of getting a bournville will be $\frac{8}{25}$

2) Probability of getting a Dairymilk silk
As there are 17 Dairymilk silk chocolates among 25,

Probability of getting a Dairymilk silk will be $\frac{17}{25}$.

3) Probability of getting Bournville and then Dairymilk silk

= $P(B)$ $\times$ $P(D)$

= $\frac{8}{25}$ $\times$ $\frac{17}{25}$

= $\frac{136}{625}$

= 0.2176

Example 3: A box has 5 white and 2 red balls. Another box has 3 white and 4 red balls. One ball is randomly selected from the first box and it is transferred to the second box. After that, one ball is randomly drawn from the second box. Find the probability that it is red.

Solution:
Let $A$ : The transferred ball is white.
$B$ : The transferred ball is red.
$C$ : The ball drawn from the second box is red.

$P(A)$ = $\frac{5}{7}$ and $P(B)$ = $\frac{2}{7}$

$P$ $(\frac{C}{A})$ = $\frac{4}{8}$ and $P$ $(\frac{C}{B})$ = $\frac{5}{8}$

Event $C$can occur along with $A$ or along with $B$. Therefore,

$P$ [Ball from second box is red] = $P (A \cap C) \cup P (B \cap C)$

= $P(A \cap C)$ + $P(B \cap C)$

= $P(A)$ . $P$ $(\frac{C}{A})$ + $P(B)$ . $P$ $(\frac{C}{B})$

= $\frac{5}{7}$ $\times$ $\frac{4}{8}$ $+$ $\frac{2}{7}$ $\times$ $\frac{5}{8}$

= $\frac{15}{28}$

= 0.5357