The probability is a extent to which an event is likely to occur by following the ratio of favourable events to the total number of outcomes. In other word probability is used where random number of outcomes may not be predicted.

There are some basic rules for probability,

**1)** The probability of an event can not be more than 1 or less than 0.

**2)** Two probability is said to be independent when outcomes does not change the probability of each other event.

There are some basic rules for probability,

A school conduct two tests of mathematics. $45 \%$ of students passed first test and $30 \%$ of students passed both tests. What is the probability of students who passed second tests ?

Given

First test passed students =

Both tests passed students =

Second test passed students =

P(second test) = $\frac{P(both\ tests\ passed\ students)}{P(First\ test\ passed\ students)}$

P(Second test) = $\frac{0.30}{0.45}$

P(Second test) = $0.66$

Hence probability of second test passed students = $0.6$

Percentage of second test passed students = $66.6 \%$

In a bag a boy has $6$ red and $8$ blue balls, he drawn $2$ balls randomly. If first ball drawn and replaced, find the probability of getting $2$ red ball ?

Given

Red balls = $6$

Blue balls = $8$

Balls drawn = $2$

Total balls = $14$

1) Find the probability of drawing first ball = $\frac{6}{14}$

As the ball is replaced hence the probability of drawing second ball = $\frac{6}{14}$

Now

Probability of drawing both red ball = $\frac{6}{14}$ $\frac{6}{14}$ = $\frac{36}{196}$ = $0.18$

A bag contains $3$ yellow and $6$ black marbles, $2$ marbles belling drawn randomly from bag. find the probability of getting $2$ yellow marbles?

Given

Yellow marbles = $3$

Black marbles = $6$

marbles drawn = $2$

Total marbles = $9$

Find the probability of drawing first yellow marbles = $\frac{3}{9}$

The probability of drawing second yellow marbles = $\frac{2}{8}$

Now

Probability of drawing both yellow marbles = $\frac{3}{9}$ $\frac{2}{8}$ = $\frac{6}{72}$ = $0.083$

From an ordinary playing card a card is drawn. If the card is a ace or spade there is a possibility to win $5$. Find the possibility to win the game?

Given

Ordinary card in a deck = $52$

Spades in a deck = $13$

P(Spades) = $\frac{13}{52}$

Aces in a deck = $4$

P(Aces) = $\frac{4}{52}$

Ordinary card has $1$ ace as spade

$P(Spades \cap Aces)$ = $\frac{1}{52}$

Hence,

The possibility of winning the game is,

$P(Spades \cup Aces)$ = $P(Spades) + P(Aces) - P(Spades \cap Aces)$

$P(Spades \cup Aces)$ = $\frac{13}{52}$ + $\frac{4}{52}$ - $\frac{1}{52}$

$P(Spades \cup Aces)$ = $\frac{16}{52}$

$P(Spades \cup Aces)$ = $\frac{4}{13}$

The possibility of winning the game is = $0.31$