Probability is a numerical measure which indicates the chance of occurrence.
For example: when a die is rolled you may get any outcome like 1, 2, 3, 4, 5 or 6.
There are three different type of approaches used in probability namely:
1) Classical approach
2) Empirical approach
3) The Axiomatic approach.
These three approaches have their own advantages and disadvantages. In this page you will learn to solve different types of problems in probability.

Conditional Problems

The conditional probability of an event will occur when another event is already occurred. Below are some problems on conditional probability.

Example 1: A box contains 7 white and 3 pink balls. Three balls are drawn at random from the box. Find the probability that both of them are white when,
1)  Drawn without replacement.
2)  Drawn with replacement.

Solution :
1) Drawn without replacement.
The balls are drawn one after the other. The balls drawn are not returned to the box in this case.
Let $A$ : First ball drawn is White.
$B$ : Second ball drawn is White.

$P(A)$ = $\frac{7}{10}$

$P$ $(\frac{B}{A})$ = $\frac{6}{9}$ = $\frac{2}{3}$, because when A has happened, the box will have only 6 white and 3 pink balls.

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

= $P(A)$ . $P$ $(\frac{B}{A})$

= $\frac{7}{10}$ $\times$ $\frac{2}{3}$

= $\frac{14}{30}$

= 0.47

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 White.
$B$ : Second ball drawn is White.

$P(A)$ = $\frac{7}{10}$
Contents of the bag remain unaltered as the first ball drawn is returned before the second draw.

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

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

= $\frac{7}{10}$ $\times$ $\frac{7}{10}$

= $\frac{49}{100}$

= 0.49

Example 2: A van has 15 packets of mangoes, among them 7 packets are Alphonso and 8 packets are kesar.
1) Find the probability of getting Kesar?
2) Find the probability of getting Alphonso?
3) Find the probability of getting Alphonso and then Kesar?

Solution :
Given:
A van has 15 packets of mangoes. 7 packets are Alphonso and 8 packets are kesar.
1) Probability of getting Kesar.
As there are 8 kesar packets of mangoes among 15,

Probability of getting a kesar packet will be $\frac{8}{15}$

2) Probability of getting Alphonso.
As there are 7 packets of Alphonso among 15,

Probability of getting a Alphonso will be $\frac{7}{15}$.

3) Probability of getting Alphonso and then Kesar.
= $P(A) \times P(K)$

= $\frac{8}{15}$ $\times$  $\frac{7}{15}$

= $\frac{56}{225}$

= 0.25

Problems based on Approaches

Let’s take a look at a couple of problems based on approaches:

Example 1:
Three unbiased coins are tossed, what is the probability of obtaining:
3)  Atleast one head
4)  Atleast two heads
5)  All tails

Solution:
Random Experiment: Three unbiased coins are tossed.
Sample space $S$ = {HHH, HHT, HTT, TTT, TTH, THH, THT, HTH} So $n$ = 8.
Probability = $\frac{Number\ of\ ways\ a\ certain\ outcome\ can\ occur}{Total\ Possible\ outcomes}$ = $\frac{m}{n}$

1) $A$ : Getting all heads
$A$ = {HHH}, i.e. $m$ = 1

$P(A)$ = $\frac{1}{8}$

2) $B$: Getting two heads, i.e. $m$ = 3
$B$ = {(HHT), (HTH), (THH)}

$P(B)$ = $\frac{3}{8}$

3) $C$: Atleast one head
$C$ = {(HHH), (HHT), (HTT),(TTH), (THH), (THT), (HTH)}, $m$ = 7

$P(C)$ = $\frac{7}{8}$

4) $D$: Getting all tails
$D$ = { TTT }, i.e. $m$ = 1

$P(D)$ = $\frac{1}{8}$

5) $E$: Getting atleast two heads
$E$ = {(HHH), (HHT), (THH), (HTH)}, i.e. $m$ = 4

$P(E)$ = $\frac{4}{8}$

Example 2: Two fair dice are rolled. Find the probability that
1) Both the dice show the number 6.
2)The sum of the numbers obtained is 10.
3) The sum is divisible by 3.

Solution: Random Experiment: Two fair dice are rolled.
$S$ = {(1,1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (3,1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (4, 1), (4, 2), (4,3), (4, 4), (4, 5), (4, 6), (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6), (6,1), (5, 2), (6, 3), (6, 4), (6, 5), (6, 6)}

So, $n$ = 36

Probability = $\frac{Number\ of\ ways\ a\ certain\ outcome\ can\ occur}{Total\ Possible\ outcomes}$ = $\frac{m}{n}$1) $A$: Both the dice show the number 6.

$P(A)$ = $\frac{1}{36}$

2) $B$: The sum of the numbers obtained is 10.
$B$ = {(1, 9), (2, 8), (3, 7), (4, 6), (5, 5), (6, 4), (7, 3), (8, 2), (9,1)}

$P(B)$ = $\frac{9}{36}$

3) $C$: The sum is divisible by 3.
$C$ = {(1, 5), (1, 2), (2, 1), (2, 4), (3, 3), (6, 3), (4, 5), (5, 4), (6, 3), (6, 6), (5, 1), (4, 2)}, $m$ = 12

$P(C)$ = $\frac{12}{36}$

Example 3: There are 17 balls, numbered from 1 to 17 in a bag. If a person selects one at random, what is the probability that the number printed on the ball will be an even number greater than 9?

Solution:
Random Experiment: Selecting a ball from a bag containing 17 balls at random.

One ball can be selected from 17 balls in $17C_{1}$ ways. $n$ = 17

$A$: Getting an even number greater than 9.

Selecting one of the following balls which are {10, 12, 14, 16}

$m$ = 4$C_{1}$

Therefore, $P(A)$ = $\frac{4}{17}$ = 0.23

Problems based on Independence

Given below are few examples based on probability where events are independent.

Example: If $P(A)$ = 0.4, $P(B)$ = 0.7 and $P(A \cup B)$ = 0.9 find $P$$(\frac{A}{B})$. Are A and B independent events?

Solution: By addition theorem
$P(A \cup B)$ = $P(A)$ + $P(B)$ - $P(A \cap B)$

Rearranging the terms

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

= 0.4 + 0.7 - 0.9

= 0.2

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

= $\frac{0.2}{0.7}$

= 0.28

Thus $P$ $(\frac{A}{B})$  = 0.28

Here $P$ $(\frac{A}{B})$ $\neq$ $P(A)$.

Hence, events $A$ and $B$ are not independent.

Example  2: One their way to work, David drives through five sets of traffic lights. The probability of each set of lights being green is 0.7. What is the probability that they are all green?

Solution :
The probability of each set of lights being green is 0.7

So $P$ [All Green] = $P$[1st green, 2nd green, 3rd green, 4th green, 5th green]

= $P$[1st green] $\times$ $P$[2nd green] $\times$ $P$[3rd green] $\times$ $P$[4th green] $\times$ $P$[5th green]

= 0.7 $\times$ 0.7 $\times$ 0.7$\times$0.7 $\times$ 0.7

= 0.16807