Probability theory is a part of mathematics that deals with the chances of occurrence of the events, means it is a measure of how likely a event will occur. That is probability is a measure of the ratio of favorable outcomes to the total number of outcomes. Let $E$ be a event and the probability of an event will be represented as $P(E)$.

$P(E)$ = $\frac{number\ of\ favorable\ outcomes}{total\ number\ of\ outcomes}$

Here outcome means one possible result of the probability, So total number of outcomes means all the possible results of the experiment, it is also called as sample space. Now favorable outcomes means the possible results which are in favorable for the event to occur.

The "$E$" written in parenthesis after '$P$' represents the particular outcome of an event  for which the probability is being calculated. For example, if you are calculating the probability of getting the head when tossing a coin, then you would write $H$ in place of "$E$", that is, you will write $P(H)$. The number of favorable outcomes in the above event will be $1$ as head  is occurred only once while tossing the coin. The number of outcomes out of the sample space when tossing a coin will be a head and a tail

The total number of possible outcomes is the total number of outcomes listed in the sample space which is $2$.

$P(H)$ = $\frac{1}{2}$

Word Problems

Example 1:

Find the probability of getting an odd number on rolling a die and a tail on tossing a coin?

Solution:

Let us first calculate all the possible outcomes that is the sample space while throwing a die.

Sample Space:

Sample Space 

Here $H$ means heads and $T$ means tails. So the sample space will be $1H, 2H, 3H, 4H, 5H, 6H, 1T, 2T, 3T, 4T, 5T$ and $6T$. So totally $12$. Now the event we need is to get an odd number on rolling a die and a tail on tossing a coin, so the favorable outcomes are $1T, 3T$ and $5T$. so if we see in the sample space there are $3$ favorable outcomes. So the favorable outcomes of getting an even number on rolling  die and a head on tossing a coin is $3$. 

Favorable Outcomes

Favorable Outcomes

$P(odd,\ tail)$ = $\frac{3}{12}$ 

                       = $\frac{1}{4}$

So the probability of getting a odd number while rolling a die and a tail while tossing a coin is $\frac{1}{4}$.
Examples 2:

Find the probability of getting a sum more than $10$ while rolling a die twice?

Solution:

Let us first calculate all the possible outcomes that is the sample space while rolling a die twice.

9th Grade Probability Examples
So the total number of outcomes is $36$. Now the event is we need is to get sum of the numbers on both the dice more than $7$, so the favorable outcomes are $(5, 6),\ (6, 5)$ and  $(6, 6)$.So if we see in the sample space there are $3$ favorable outcomes. So the favorable outcomes of getting sum of the numbers on both the die more than $10$ while rolling a die twice are $3$.

9th Grade Probability Example
 
We know that,

$P(E)$ = $\frac{number\ of\ favorable\ outcomes}{total\ number\ of\ outcomes}$

Here the event $E$ is getting an sum of more than $10$ while rolling two die, so probability will be equal to

P(same number) = $\frac{3}{36}$

P(same number) = $\frac{1}{12}$

So the probability of getting a sum equal to $7$ while rolling two dice is equal to $\frac{1}{12}$ or $0.083$ or $8.33 \%$
Example 3:

Find the probability of drawing a king from the first draw and a number card from the second draw with replacement?
Solution:

In $52$ deck of cards there are $4$ Kings and $40$ number cards. As it is with replacement these events are independent and hence the sample space will be same for both events. So first calculate probability of both the events individually. Here sample space is $52$ as there are $52$ cards in a deck of cards.

Probability of drawing a king from deck of cards be $P(K)$, then

$P(K)$ = $\frac{number\ of\ kings\ in\ deck\ of\ cards}{total\ number\ of\ cards}$

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

$P(K)$ = $\frac{1}{13}$

Probability of drawing a number card from deck of cards be $P(N)$, then 

$P(N)$ = $\frac{number\ of\ number\ cards\ in\ deck\ of\ cards}{total\ number\ of\ cards}$

$P(N)$ = $\frac{36}{52}$

$P(N)$ = $\frac{9}{13}$

Now the probability of these two events is

$P(K \cap N)$ = $P(K) \times P(N)$

= $\frac{1}{13}$ $\times$ $\frac{9}{13}$

= $\frac{9}{169}$

The probability of drawing a king from the first draw and a number card from the second draw with replacement is $\frac{9}{169}$.
Example 4: 

Two cards are chosen at random from a deck of cards without replacement. What is the probability that all of them are Face cards?

Solution:

In $52$ deck of cards there are $12$ face cards. As it is without replacement these events are dependent and hence the sample space will be deducted by $1$ as we are drawing one card at a time. 

Probability of drawing a face card from deck of cards in first draw be $P(F1)$, then 

$P(F1)$ = $\frac{number\ of\ face\ cards\ in\ first\ draw}{total\ number\ of\ cards}$

$P(F1)$ = $\frac{12}{52}$

$P(F1)$ = $\frac{3}{13}$

Probability of drawing a face card from deck of cards in second draw be $P(F2)$, then 

Here the number of face cards reduces to $11$ as already $1$ face card is drawn and the total number of cards is reduced to $51$ as one card is removed from deck of cards.

$P(F2)$ = $\frac{number\ of\ face\ cards\ in\ second\ draw}{total\ number\ of\ cards}$

$P(F2)$ = $\frac{11}{51}$

$P(F2)$ = $\frac{11}{51}$


Now the probability of these events is

$P(F1 \cap F2)$ = $P(F1) \times$ $P$ $(\frac{F1}{F2})$

$P(F1 \cap F2)$ = $\frac{1}{13}$ $\times$ $\frac{11}{51}$

$P(F1 \cap F2)$ = $\frac{11}{663}$

The probability of drawing two number $3$ cards from a deck of cards without replacement is $\frac{1}{221}$.
Examples 5: 

A bag contains $5$ yellow marbles, $16$ green marbles, and $2$ red marbles. What is the probability of drawing a red marble in first draw and a yellow marble in the second draw without replacement? 

Solution:

Given bag contains $5$ yellow marbles, $16$ green marbles, and $2$ red marbles. That is totally it have $23$ marbles $(5 + 16 + 2$ = $23)$.  So sample space is $23$. As it is without replacement these events are dependent and hence the sample space will be deducted by $1$ as we are drawing one marble at a time. 

Probability of drawing a red marble in first draw be $P(R)$, then 

$P(R)$ = $\frac{number\ of\ red\ marbles\ in\ the\ bag}{total\ number\ of\ marbles\ in\ the\ bag}$

$P(R)$ = $\frac{2}{23}$

Probability of drawing a yellow marble in second draw be $P(Y)$, then 

Here the total number of marbles is reduced to $22$ as one red marble is removed from the bag in the first draw.

$P(Y)$ = $\frac{number\ of\ yellow\ marbles\ in\ the\ bag}{total\ number\ of\ marbles\ in\ the\ bag}$

$P(Y)$ = $\frac{5}{22}$

$P(Y)$ = $\frac{5}{22}$

Now the probability of these events is

$P(R \cap Y)$ = $P(R) \times P$ $(\frac{R}{Y})$

$P(R \cap Y)$ = $\frac{2}{23}$ $\times$ $\frac{5}{22}$

$P(R \cap Y)$ = $\frac{10}{506}$ 

$P(R \cap Y)$ = $\frac{5}{253}$

The probability of drawing red marble in first draw and a yellow marble in the second draw from the given bag of marbles without replacement is $\frac{5}{253}$.