Probability is the study of how likely an event will occur.  Many events can’t be predicted with total certainty. The best we can say is how likely they are to happen, using the idea of probability.

For example, when a coin is tossed, there are two possible outcomes: Heads (H) or Tails(T)

By Definition, Probability is the ratio of favorable outcomes (the number of ways an event can happen) to total possible outcomes. 

The probability of an event A occurring is:
\(
P(A)=\frac{\text { The number of ways for } A \text { to occur }}{\text { Total number of possible outcomes }}
\)
If we take the example of a coin total possible outcomes = 2 (Head or Tail).

So, the probability of flipping heads with a fair coin H = 1/2. Similarly, the probability of flipping tails with a fair coin T = 1/2. Therefore the Probability of flipping H or T in a fair coin toss is \(50 \%\)

Example 1: Find the probability of rolling a 3 with a die.

Solution:

Number of ways it can happen: 1 (there is only 1 face with a “3” on it)
Total number of outcomes: 6 (there are 6 faces altogether).
So the probability, \(P(3)=\frac{\text { Number of ways to roll } 3}{\text { Total number of outcomes }}=\frac{1}{6}\)

Example 2: There are 5 marbles in a bag: 4 are blue, and 1 is red. What is the probability that a blue marble gets picked?

Solution:

Number of ways it can happen: 4 (there are 4 blues)
Total number of outcomes: 5 (there are 5 marbles in total)
So the probability of picking a blue marble \(P(Blue)=\frac{4}{5}=0.8\)

Example 3: There are 6 pillows in a bed, 3 are red, 2 are yellow and 1 is blue. What is the probability of picking a yellow pillow?

Solution:

Number of ways it can happen: 2 (there are 2 yellow pillows)
Total number of outcomes: 6 (there are 6 pillows in total)
So the probability of picking a yellow pillow \(P(Yellow)=\frac{2}{6}=\frac{1}{3}\)

Example 4: There is a container full of coloured bottles, red, blue, green, and orange. Some of the bottles are picked out and displaced. Sumit did this 1000 times and got the following results:

• Number of blue bottles picked out: 300
• Number of red bottles: 200
• Number of green bottles: 450
• Number of orange bottles: 50

What is the probability that Sumit will pick a green bottle? If there are 100 bottles in the container, how many of them are likely to be green?

Solution:

For every 1000 bottles picked out, 450 bottles are green. Number of ways it can happen is 450. Total number of outcomes = 1000.
Therefore, \(P(Green)=450 / 1000=0.45\)

The experiment implies that 450 out of 1000 bottles are green. Therefore, out of 100 bottles, 45 are green.

Probability Terms and Definition

Some words have special meanings in Probability. Some of the important probability terms generally used are discussed here:

Experiment or Trial: a repeatable procedure with a set of possible results.

Example: Throwing a dice.

We can throw the dice again and again, so it is repeatable. 
\(\text { The set of possible results from any single throw is }\{1,2,3,4,5,6\}\)

             

Other examples are the tossing of a coin and selecting a card from a deck of cards.

Outcome: Possible result of an experiment or trial.

Example: “4” is one of the outcomes of a throw of a die. Similarly, T (tail) or H(head) is a possible outcome when a coin is tossed.

           

Sample Space: All the possible outcomes of an experiment. 

Example: choosing a card from a deck. There are 52 cards in a deck (not including Jokers)

So the Sample Space is all 52 possible cards: {Ace of Hearts, 2 of Hearts, etc… }

             

Other examples are, Tossing a coin, Sample Space \((\mathrm{S})=\{\mathrm{H}, \mathrm{T}\}\)
Rolling a die, Sample Space \((S)=\{1,2,3,4,5,6\}\)

Another example is throwing a dice. There are 6 different sample points in that sample space.

\(\text { In rolling a dice, we have, } S=\{1,2,3,4,5,6\} \text {. }\) \(\text { If two coins are tossed, then } \mathrm{S}=\{\mathrm{HH}, \mathrm{HT}, \mathrm{TH}, \mathrm{TT}\} \text {. }\)

             

Note: Sample Space: When we perform an experiment, then the set S of all possible outcomes is called the Sample Space.

Event: One or more outcomes of an experiment.

Example: An event can be just one outcome:

• Getting a Tail when tossing a coin
• Rolling a “5” in a dice.

An event can also include more than one outcome:

• Choosing a “King” from a deck of cards (any of the 4 Kings)
• Rolling an “even number” (2, 4, 6, or 8)

Complimentary Event: The non-happening events. The complement of an event \(A\) is the event, not \(A\) (or \(A^{\prime}\) ).

Example: In a standard 52-card deck, A \(=\) Draw a heart, then \(\mathrm{A}^{\prime}=\) Don’t draw a heart.

Impossible Event: The event cannot happen. 

Example: In tossing a coin, impossible to get both head and tail at the same time.

Equally Likely Events

When the events have the same probability of happening, then they are called equally likely events.

Example: If you throw a dice, then the probability of getting 1 is 1/6. Similarly, the probability of getting all the numbers from 2,3,4,5, and 6, one at a time is 1/6. 

The probability of an event not happening

If we know the probability of an event occurring, we can also work out the probability of it not happening. For example, if we wanted to work out the probability of rolling a number between 1 and 6, we can write the following:

\(P(\) rolling \(1,2,3,4,5\), or 6\()=\frac{1+1+1+1+1+1}{6}=1\)
Intuitively, the event rolling between 1 and 6 is a certainty since these are the only options available.

The fact that all probabilities must add up to 1 is useful in working out the probability of something not happening. For example, the probability of not rolling a three is the same as the probability of rolling either 1, 2, 4, 5, or 6, which, using the formula, is
\(
P(\text { not rolling } 3)=\frac{1+1+1+1+1}{6}=\frac{5}{6} \text {. }
\)
Alternatively, we could have subtracted the probability of failure from 1 :
\(
P(\text { not rolling } 3)=1-P(\text { rolling } 3)=1-\frac{1}{6}=\frac{5}{6} \text {. }
\)

Definition

The probability of an event, A, not happening is:
\(
P(\text { not } A)=1-P(A) \text {. }
\)
Using set notation:
\(
P\left(A^{\prime}\right)=1-P(A) \text {. }
\)

Therefore, we can say \(P(A)+P\left(A^{\prime}\right)=1\)

Important Points on Probability

• \(\mathrm{P}(\mathrm{S})=1\)
• \(0 \leq P\) (E) \(\leq 1\)
• \(\mathrm{P}(\phi)=0\)
• For any events \(A\) and \(B\), we have :
  \(
  P(A \cup B)=P(A)+P(B)-P(A \cap B)
  \)
• If \(A^{\prime}\) denotes (not-A), then \(\mathrm{P}(A^{\prime})=1-\mathrm{P}(\mathrm{A})\).

Example 5: What is the probability of rolling a 1, 2, 3, or 6?

Solution:

Since this is equivalent to asking what the probability of not rolling a 4 or 5 , we can do the following:
\(P(\) rolling 4 or 5\()=\frac{1+1}{6}=\frac{1}{3}\)
Therefore, using the formula \(P(\operatorname{not} A)=1-P(A)\) :
\(P(\) rolling \(1,2,3\) or 6\()=1-P(\) rolling 4 or 5\()=1-\frac{1}{3}=\frac{2}{3}\)

Example 6: There are 10 balls in a bag out of which 3 are black, 2 are red, 1 is blue, 2 are pink, and 2 are purple. Let \(X\) be the event of selecting a primary color. Find \(\mathrm{P}\left(\mathrm{X}^{\prime}\right)\).

Solution:

\(X=\{2\) red, 1 blue \(\}\)
Total balls \(=10\)
Number of favorable outcomes \(=3\)
\(
\mathrm{P}(\mathrm{X})=3 / 10
\)
Using the rule of complementary events, \(P\left(A^{\prime}\right)=1-P(A)\)
\(
\mathrm{P}\left(\mathrm{X}^{\prime}\right)=1-(3 / 10)=7 / 10
\)

Example 7: A random number is chosen from 1 to 50. Calculate the probability of not choosing a perfect square.

Solution:

Let Z’ be the event of choosing a perfect square. The sample space is given as follows:
\(
Z^{\prime}=\{1,4,9,16,25,36,49\}
\)
Total number of outcomes \(=50\)
Favorable outcomes \(=7\)
\(
\begin{aligned}
& P\left(Z^{\prime}\right)=7 / 50 . \\
& P(Z)=1-(7 / 50) \\
& =43 / 50
\end{aligned}
\)

Example 8: Find the probability of ‘getting 3 on rolling a dice.

Solution:

Sample Space \(=S=\{1,2,3,4,5,6\}\)
Total number of outcomes \(=n(S)=6\)
Let \(\mathrm{A}\) be the event of getting 3.
Number of favourable outcomes \(=n(A)=1\)
Probability, \({P}({A})=\mathrm{n}({A}) / \mathrm{n}({S})=1 / 6\)
Hence, \({P}\) (getting 3 on rolling a dice \()=1 / 6\)

Example 9: A vessel contains 4 blue balls, 5 red balls, and 11 white balls. If three balls are drawn from the vessel at random, what is the probability that the first ball is red, the second ball is blue, and the third ball is white?

Solution:

Given, the Total number of balls or possibilities is 20.
The probability to get the first ball is red (out of 5 red balls any one red ball can be picked) \(5 / 20\).
Since we have drawn a ball for the first event to occur, then the number of possibilities left for the second event to occur is \(20-1=19\).
Hence, the probability of getting the second ball as blue is (any blue ball can be picked from 4 blue balls) 4/19.
Again with the first and second event occurring, the number of possibilities left for the third event to occur is \(19-1=18\).
And the probability of the third ball is white is 11/18.
Therefore, the probability is \(5 / 20 \times 4 / 19 \times 11 / 18=44 / 1368=0.032\).
Or we can express it as: \(P=3.2 \%\).

Example 10: Two dice are rolled, find the probability that the sum is: (i) equal to 1 (ii) equal to 4 (iii) less than 13.

Solution:

To find the probability that the sum is equal to 1 (each dice has 6 faces (1,2,3,4,5,and 6) we have to first determine the sample space \(S\) of two dice as shown below.
\(
\begin{aligned}
& S=\{(1,1),(1,2),(1,3),(1,4),(1,5),(1,6) \\
& (2,1),(2,2),(2,3),(2,4),(2,5),(2,6) \\
& (3,1),(3,2),(3,3),(3,4),(3,5),(3,6) \\
& (4,1),(4,2),(4,3),(4,4),(4,5),(4,6) \\
& (5,1),(5,2),(5,3),(5,4),(5,5),(5,6) \\
& (6,1),(6,2),(6,3),(6,4),(6,5),(6,6)\} \\
& \text { So, } n(S)=36
\end{aligned}
\)

(i) Let \(E\) be the event “sum equal to 1”. Since, there are no outcomes which where a sum is equal to 1, hence,
\(
\mathrm{P}(\mathrm{E})=\mathrm{n}(\mathrm{E}) / \mathrm{n}(\mathrm{S})=0 / 36=0
\)
(ii) Let \(A/\) be the event of getting the sum of numbers on dice equal to 4.
Three possible outcomes give a sum equal to 4 they are:
\(A=\{(1,3),(2,2),(3,1)\}\)
\(n(A)=3\)
Hence, \(P(A)=n(A) / n(S)=3 / 36=1 / 12\)
(iii) Let \(B\) be the event of getting the sum of numbers on dice is less than 13.
From the sample space, we can see all possible outcomes for the event B, which gives a sum less than B. Like:
\(
(1,1) \text { or }(1,6) \text { or }(2,6) \text { or }(6,6) \text {. }
\)
So you can see the limit of an event to occur is when both dies have number 6 , i.e. \((6,6)\).
Thus, \(n(B)=36\)
Hence,
\(
P(B)=n(B) / n(S)=36 / 36=1
\)

Example 11: A fair die and a coin were rolled once. What is the probability of; (a) having an even number? (b) having either an even or a prime number.

Solution:

(a) having an even number;
To find the probability of an even number, We know we have three even numbers in a fair die, that is 2,4 , and 6 . Therefore, the probability of an even number becomes;
\(
P(\text { even })=\frac{3}{6}
\)
(b) having either an even or a prime number;
In the question (a), we found the probability of finding an even number as;
\(
P(\text { even })=\frac{1}{2}
\)
Next, we need to find the probability of a prime number occurring.
We know we have three prime numbers in a fair die, that is 2, 3, and 5. Therefore, the probability of a prime number becomes;
\(
P(\text { prime })=\frac{3}{6}=\frac{1}{2}
\)
Therefore the probability of an even or a prime number is;
\(
P(\text { even or prime })=\frac{1}{2}+\frac{1}{2}=1
\)

Example 12: Two unbiased coins are tossed. What is the probability of getting at most one head?

Solution:

Here \(S=\{\mathrm{HH}, \mathrm{HT}, \mathrm{TH}, \mathrm{TT}\}\).
Let \(E=\) event of getting at most one head.
\(
\begin{aligned}
& \therefore \quad E=\{\mathrm{HH}, \mathrm{HT}, \mathrm{TH}\} \\
& \therefore \quad P(E)=\frac{n(E)}{n(S)}=\frac{3}{4}
\end{aligned}
\)

Example 13: An unbiased die is tossed. Find the probability of getting a multiple of 3.

Solution:

Here \(S=\{1,2,3,4,5,6\}\)
Let \(E\) be the event of getting a multiple of 3 .
Then, \(E=\{3,6\}\).
\(
\therefore \quad P(E)=\frac{n(E)}{n(S)}=\frac{2}{6}=\frac{1}{3} .
\)

Example 14: In a simultaneous throw of a pair of dice, find the probability of getting a total more than 7.

Solution:

Here, \(n(\mathrm{~S})=(6 \times 6)=36\)
Let \(E \quad=\) Event of getting a total more than 7
\(=\{(2,6),(3,5),(3,6),(4,4),(4,5),(4,6),(5,3),(5,4),(5,5),(5,6),(6,2),(6,3),(6,4),(6,5),(6,6)\}\).
\(\therefore \quad P(E)=\frac{n(E)}{n(S)}=\frac{15}{36}=\frac{5}{12}\).

Example 15: Two dice are thrown together. What is the probability that the sum of the numbers on the two faces is divisible by 4 or 6?

Solution:

Clearly, \(n(\mathrm{~S})=6 \times 6=36\)
Let \(\mathrm{E}\) be the event that the sum of the numbers on the two faces is divisible by 4 or 6.
Then \(E=\{(1,3),(1,5),(2,2),(2,4),(2,6),(3,1),(3,3),(3,5),(4,2),(4,4),(5,1),(5,3),(6,2),(6,6)\}\)
\(\therefore \quad n(E)=14\)
Hence, \(P(E)=\frac{n(E)}{n(S)}=\frac{14}{36}=\frac{7}{18}\).

Example 16: A bag contains 6 white and 4 black balls. Two balls are drawn at random. Find the probability that they are of the same colour.

Solution:

Let \(S\) be the sample space. Then, \(n(s)=\) Number of ways of drawing 2 balls out of \((6+4)={ }^{10} C_2=\frac{(10 \times 9)}{(2 \times 1)}=45\). Let \(E=\) Event of getting both balls of the same colour. Then, \(n(E)=\) Number of ways of drawing ( 2 balls out of 6 ) or ( 2 balls out of 4 )
\(
\begin{aligned}
& =\left({ }^6 C_2+{ }^4 C_2\right)=\frac{(6 \times 5)}{(2 \times 1)}+\frac{(4 \times 3)}{(2 \times 1)}=(15+6)=21 . \\
\therefore \quad \mathrm{P}(\mathrm{E}) & =\frac{n(E)}{n(S)}=\frac{21}{45}=\frac{7}{15} .
\end{aligned}
\)

Example 17: Two cards are drawn at random from a pack of 52 cards. What is the probability that either both are black or both are queens?

Solution:

We have \(n(s)={ }^{52} C_2=\frac{(52 \times 51)}{(2 \times 1)}=1326\)
Let \(A=\) event of getting both black cards;
\(B=\) event of getting both queens.
\(\therefore \quad A \cap B=\) event of getting queens of black cards.
\(\therefore \quad n(A)={ }^{26} C_2=\frac{(26 \times 25)}{(2 \times 1)}=325, n(B)={ }^4 C_2=\frac{(4 \times 3)}{(2 \times 1)}=6\) and \(n(A \cap B)={ }^2 C_2=1\).
\(\therefore \quad P(A)=\frac{n(A)}{n(S)}=\frac{325}{1326} ; P(B)=\frac{n(B)}{n(S)}=\frac{6}{1326}\) and \(P(A \cap B)=\frac{n(A \cap B)}{n(S)}=\frac{1}{1326}\).
\(\therefore \quad P(A \cup B)=P(A)+P(B)-P(A \cap B)=\left(\frac{325}{1326}+\frac{6}{1326}-\frac{1}{1326}\right)=\frac{330}{1326}=\frac{55}{221}\).

Example 18: A positive integer less than 100 is selected at random. What is the probability that the integer is odd?

Solution:

There are 99 integers less than 100 and 50 of them are odd. So the probability of selecting an odd integer is 50/99.

Example 19: What is the probability of rolling a 6 with a pair of standard six-faced dice?

Solution:

There are five ways to roll a 6 {(1,5),(2,4),(3,3),(4,2), and (5,1)}. There are 6 x 6=36 possible outcomes when a pair of dice is rolled. The probability of rolling a 6 with a pair of dice is 5/36.

Example 20:  All of the arrangements of the letters in the word ALGEBRA are written on a list, and one of the arrangements is selected at random. What is the probability that the selected arrangement contains a double-A?

Solution:

Tie the two A’s together and pretend they are one letter. Now, we have something like AALEGBR. Since AA is one letter, this can be arranged in 6!=720 ways.
There are 7!/2 arrangements of the word ALGEBRA.
Explanation: Just assume that 2 A’s are different. Let’s call one of them as \(A_1\) and the other \(A_2\). Now we have 7 letters and the total number of words that can be formed with 7 letters is \(P(7,7)=7!=5040\) words. However, we overcounted. \(A_1LGEBRA_2\) is the same as \(A_2LGEBRA_1\). Similarly \(A_1A_2LGEBR\) is the same as \(A_2A_1LGEBR\). To eliminate the extra cases, we need to divide by the ways that the A’s can be arranged, which in this case is just 2!=2.
\(P(7,7)/2=2520\) words.

So, the probability is 720/2520=2/7