Probability

Probability defines the likelihood of occurrence of an event. There are many real-life situations in which we may have to predict the outcome of an event. We may be sure or not sure of the results of an event. In such cases, we say that there is a probability of this event to occur. It is to be known that probability and statistics are related to each other, where the first one deals with predicting future events and statistics deals with data handling, data representation.

Cards, dice

In this mini-lesson, we will discuss about basics of probability, terms used in probability, probability formula, conditional probability, probability distribution, probability density function, law of total probability, Bayes' theorem.

Lesson Plan

Basics of Probability

Probability can be defined as the ratio of the number of favorable outcomes to the total number of outcomes of an event. 

For example, let us consider the following statement.

Will it rain today? 

The answer to this question is either "Yes" or "No." There is a likelihood to rain or not rain.

This is called probability.

Some important terms used in probability are as discussed below.

Experiment

A trial or an operation conducted to produce an outcome is called an experiment.

Sample Space 

All the possible outcomes of an experiment together constitute a sample space. For example, the sample space of tossing a coin are head and tail.

Favorable Outcome

An event that has produced the desired result or expected event is called a favorable outcome. For example, when we roll two dice, the possibility of getting number 4 is (1,3), (2,2), and (3,1).

Trial

A trial denotes doing a random experiment.

Random Experiment

An experiment in which we know all the possible outcomes but are not the exact income is called a random experiment. For example, when we toss a coin, we know that we would get a head or tail, but we are not sure of which one will appear.

Event

The total number of outcomes of a random experiment is called an event.

Equally Likely Events

Events that have the same chances or probability of occurring are called equally likely events. The outcome of one event is independent of the other.

For example, when we toss two coins, there is a chance that we get head while tossing both the coins. 

Exhaustive Events

When the set of all outcomes of an experiment is equal to the sample space, we call it an exhaustive event.

Mutually Exclusive Events

Events that are not happening simultaneously are called mutually exclusive events. For example, the climate can be either hot or cold. We cannot experience the same weather simultaneously.

Probability of an Event Formula

\(Probability\: of\: an\: event\:P(E)\:\)=\(\: \dfrac{Number\: of\: favorable\: outcomes}{Sample Space}\)

Tossing a Coin 

Let us now look into the probability of tossing a coin and rolling a dice.

Paul and his friend Joe wanted to play a game. They both wanted to take the first turn to play.  Paul's mom asked Paul and John to decide on who wanted the head of the coin and who wanted the tail of the coin. Fortunately, they both decided on different options. John wanted a head and Paul wanted a tail. She then asked one of them to toss a coin and if head occurs John can take the first turn to play and if tail occurs Paul would take the first turn to play. They happily agreed on this method and enjoyed playing their game. 

Tossing a coin

We can also see coining tossing in games such as cricket to decide on who should bowl or bat first.

This method is used because the chances of getting a head and tail are equal.

This means there is a 50% chance of getting a head and a 50% chance of getting a tail.

Tossing a Coin Probability Formula

While tossing a coin, there are two possible outcomes. One is head and the other is tail.

\(\begin{align} Total\: number\: of\: possible\: outcomes\: =\: 2\: (Head\: and\: Tail)\end{align}\)

\(\begin{align} Number\: of\: favorable\: outcome\: to\: get\: a\: head\:=\: 1\:  (Head)\end{align}\)

\(\begin{align} Number\: of\: favorable\: outcome\: to\: get\: a\: tail\:=\: 1\:  (Tail)\end{align}\)

Therefore,

\(\begin{align} Probability\: of \:getting\: a\: head\: =\: \dfrac{Number\: of\: favorable\: outcome\: to\: get\: a\: head}{Total\: number\: of\: possible\: outcomes} \end{align}\)

\(\begin{align} Probability\: of \:getting\: a\: tail\: =\: \dfrac{Number\: of\: favorable\: outcome\: to\: get\: a\: tail}{Total\: number\: of\: possible\: outcomes} \end{align}\)

So, the probability of getting a head = \(\begin{align} \dfrac{1}{2} \end{align}\)

the probability of getting a tail = \(\begin{align} \dfrac{1}{2} \end{align}\)

Please note that when we toss two or more coins together the probability of getting a head or tail for each coin may differ.


Rolling a Die

Rolling a Single Die 

Dice finds its use in playing a lot of board games to decide how many squares are we supposed to move. Similar to tossing a coin, rolling a die also has a probability. 

The set of possible outcomes when we roll a die are \(\begin{align} \{{1,2,3,4,5,6}\} \end{align}\). We may get any number on the face of a die when we roll it.

For example,

The probability of getting a desired number while rolling a die is,

\(\begin{align} Probability\: of \:getting\: the\: desired\:number \: =\: \dfrac{Number\: of\: favorable\: outcome\: to\: get\: the\:desired\:number}{Total\: number\: of\: possible\: outcomes} \end{align}\)

Probability of getting a 3 while rolling a die = \(\begin{align} \dfrac{1}{6} \end{align}\) (Since there is a probability to get number \(\begin{align} 3\end{align}\) once)

Rolling Two Dice

Sometimes we roll two dice together.

Probability of throwing a dice

All the probabilities of rolling two dice together are presented in the table below.

Probability of rolling two dice


Rolling Three Or More Dice

As we see, when we roll a single die, there are 6 possibilities. When we roll two dice, there are 36 possibilities. When we roll 3 dice we get 216 possibilities. So a general formula to represent this is,

\(\begin{align} Probability\: of\: rolling\: a\:die\: 'n' \: times\: =\: 6^n\end{align}\)

Probability of Drawing Cards

A deck containing 52 cards are grouped into four suits of clubs, diamonds, hearts, and spades. Now let us discuss the probability of drawing cards from a pack.

The symbols on the cards are shown below. Spades and clubs are black cards. Hearts and diamonds are red cards.

Cards in a pack

In each suit there is an ace, king, queen, jack, 10, 9, 8, 7, 6, 5, 4, 3, 2. King, queen, and jack are called face cards. We can apply the same formula of probability to find the probability of drawing a single card or two or more cards. 

For example, the probability of drawing a ace spade from a pack of 52 cards is  \(\begin{align}\dfrac{1}{52}\end{align}\).

We will look into more examples in the Solved Examples section.


Basic Theorems on Probability

Before getting to the theorems of probability, let us discuss a few important points.

The probability value ranges from 0 to 1, where 0 refers to the possibility of occurrence of an event and 1 denotes impossibility. 

The sum of probability of all possible events is 1

Theorem 1

\(\begin{align}P(A) = P(1-A')\end{align}\)

Theorem 2

\(\begin{align}P(\phi) =0\end{align}\)

Theorem 3

If there are two events A and B and A is a subset of B,  then,

\(\begin{align}P(A) \leq P(B)\end{align}\)

Theorem 4

The probability of any event is always less than or equal to 1

\(\begin{align}P(A) \leq 1\end{align}\)

Theorem 5

If there are two events A and B, then

\(\begin{align}P(A\cup B) = P(A) + P(B) - P(A\cap B)\end{align}\)

Conditional Probability or Bayes' Theorem

Bayes' theorem describes the probability of an event based on the condition of occurrence of other events. It is also called conditional probability.

For example, let us assume that there are three bags with each bag containing some blue, green, and yellow balls. What is the probability of picking a yellow ball from the third bag? Since there are also blue and green coloured balls, we can arrive at the probability based on these conditions also. Such a probability is called a conditional probability.

The formula for Bayes' theorem is 

\(\begin{align}P(A|B) = \dfrac{P(A) P(B|A)} {P(B)}\end{align}\)

where \(\begin{align}P(A|B) \end{align}\) denotes how often event A happens on a condition that B happens.

where \(\begin{align}P(B|A) \end{align}\) denotes how often event B happens on a condition that A happens.

\(\begin{align}P(A) \end{align}\) the likelihood of occurrence of event A.

\(\begin{align}P(B) \end{align}\) the likelihood of occurrence of event B.

Law of Total Probability

The Law of total probability states that when there is not enough information to find the probability of event A, then we take a related event B and use it to calculate the probability of A. 

For example, consider the figure below.

Total probability rule

In the figure, it is clear that the probability of A cannot be found alone. The probability of A depends on the occurrence of events  B and C. But it can be found collectively with the occurrence of condition B and condition C.

The formula for law of probability is

\(\begin{align}P(A) = \sum_{n} \: P(A \cap B_n) \end{align}\)

Where,

\(\begin{align}n \end{align}\) is the total number of events.

\(\begin{align}B_n \end{align}\) is the total number of distinct events.

 
important notes to remember
Important Notes
  1. Probability is a measure of how likely an event is to happen.
  2. Probability is represented as a fraction and always lies between 0 and 1
  3. An event can be defined as a subset of sample space.
  4. The outcome of throwing a coin is a head or a tail and the outcome of throwing a dice is 1, 2, 3, 4, 5, or 6
  5. A random experiment cannot predict the exact outcomes but only some probable outcomes.

Solved Examples

Example 1

 

 

A coin is tossed 3 times. Write down the set of sample space.

Solution

We get a head or a tail while we toss a coin. Let us denote head by 'H' and tail by 'T".

Therefore the sample space = \(\begin{align}\{{HHH, HHT, HTH, THH, TTH, THT, HTT, TTT}\}\end{align}\)

Sample space of tossing 3 coins is \(\begin{align}\{{HHH, HHT, HTH, THH, TTH, THT, HTT, TTT}\}\end{align}\)
Example 2

 

 

What is the probability of getting a sum of 10 when two dice are thrown?

Solution

There are totally 36 possibilities when we throw two dice. 

Desired outcome is 10. To get 10, we can have the following favorable outcomes.

\(\begin{align}(4,6),(6,4),(5,5)\}\end{align}\) . There are three favorable outcomes. 

\(\begin{align} Probability\: of\: an\: event\:P(E)\:=\: \dfrac{Number\: of\: favorable\: outcomes}{Sample \: space}\end{align}\)

Probability of getting number \(\begin{align} 10\:=\: \dfrac{3}{36}\end{align}\)

which simplifies to 

\(\begin{align} \dfrac{1}{12}\end{align}\)

Probability of getting a sum of 10 when rolling two dice is \(\begin{align} \dfrac{1}{12}\end{align}\)
Example 3

 

 

In a bag, there are 6 blue balls and 8 yellow balls. Two balls are selected randomly without replacement. Find the probability of getting a blue ball and a yellow ball.

Solution

Let us assume the probability of drawing a blue ball to be \(\begin{align}P(B)\end{align}\)

Number of favorable outcomes to get a blue ball  = \(\begin{align}6\end{align}\)

Number of favorable outcomes to get a yellow ball =  \(\begin{align}8\end{align}\)

Sample space =  \(\begin{align}14\end{align}\)

and the probability of drawing a green ball to be \(\begin{align}P(Y)\end{align}\)

\(\begin{align}P(B) = \dfrac{6}{14}\end{align}\) 

\(\begin{align}P(Y) = \dfrac{8}{14}\end{align}\) 

Therefore the probability of drawing a blue and a yellow ball from the bag is, 

\(\begin{align}P(B) \times P(Y)\end{align}\)

\(\begin{align}\dfrac{6}{14} \times \dfrac{8}{14}\end{align}\)

\(\begin{align}\dfrac{48}{196} \end{align}\)

which simplifies to,

\(\begin{align}\dfrac{12}{49} \end{align}\)

Probability of drawing a blue and a yellow ball from a bag of 14 balls is \(\begin{align}\dfrac{12}{49} \end{align}\)
Example 4

 

 

Find the probability of the following events.

a)  "4" of diamonds

b) Queen

c) Black King

d) A heart card

e) A king or a queen

f) A non-face card

g) neither a diamond or a black king

Solution

a) "4" of diamonds

\(\begin{align} Probability\: of\: an\: event\:P(E)\:=\: \dfrac{Number\: of\: favorable\: outcomes}{Sample \:space}\end{align}\)

Number of favorable outcomes to get a '4' card is 1 out of 52 cards

So, \(\begin{align} P(E)\:=\: \dfrac{1}{52}\end{align}\)

b) Queen

\(\begin{align} Probability\: of\: an\: event\:P(E)\:=\: \dfrac{Number\: of\: favorable\: outcomes}{Sample \: space}\end{align}\)

Number of favorable outcomes to get a queen card is 4, since we have 4 queen cards.

So, \(\begin{align} P(E)\:=\: \dfrac{4}{52}\end{align}\)

which simplifies to \(\begin{align} P(E)\:=\: \dfrac{1}{13}\end{align}\)

c) A black king

\(\begin{align} Probability\: of\: an\: event\:P(E)\:=\: \dfrac{Number\: of\: favorable\: outcomes}{Sample \: space}\end{align}\)

Number of favorable outcomes to get a black king is 2, since we have 2 black king cards.

So, \(\begin{align} P(E)\:=\: \dfrac{2}{52}\end{align}\)

which simplifies to \(\begin{align} P(E)\:=\: \dfrac{1}{26}\end{align}\)

d) A heart card

\(\begin{align} Probability\: of\: an\: event\:P(E)\:=\: \dfrac{Number\: of\: favorable\: outcomes}{Sample \: space}\end{align}\)

Number of favorable outcomes to get a heart card is 13, since we have 13 heart cards.

So, \(\begin{align} P(E)\:=\: \dfrac{13}{52}\end{align}\)

which simplifies to \(\begin{align} P(E)\:=\: \dfrac{1}{4}\end{align}\)

e) A king or a queen

\(\begin{align} Probability\: of\: an\: event\:P(E)\:=\: \dfrac{Number\: of\: favorable\: outcomes}{Sample \: space}\end{align}\)

Number of favorable outcomes to get a king is 4, since we have 4 king cards.

The number of favorable outcomes to get a queen is 4, since we have 4 queen cards.

So, 

\(\begin{align} P(E)\:&=\: \dfrac{4}{52} + \dfrac{4}{52} \\\ P(E)\:&=\: \dfrac{8}{52}\end{align}\) 

which simplifies to \(\begin{align}  \dfrac{2}{13}\end{align}\)

f) A non-face card

\(\begin{align} Probability\: of\: an\: event\:P(E)\:=\: \dfrac{Number\: of\: favorable\: outcomes}{Sample \: space}\end{align}\)

Total number of face cards is  \(\begin{align} 12\end{align}\)

Total number of non-face cards =  \(\begin{align} 52 - 12\end{align}\)

Total number of non-face cards =  \(\begin{align} 40\end{align}\)

So, \(\begin{align} P(E)\:=\: \dfrac{40}{52} \end{align}\)

which simplifies to \(\begin{align} P(E)\:=\: \dfrac{10}{13}\end{align}\)

g) Neither a diamond nor a black king

Total number of diamond cards = \(\begin{align} 13\end{align}\)

Total number of non-diamond cards = \(\begin{align} 52-13 = 39 \end{align}\)

We have excluded the diamond cards now.

Total number of red kings is \(\begin{align} 2\end{align}\)

So, the possible outcome of drawing neither a diamond or a black king = \(\begin{align} 39 -1 = 38\end{align}\)

We have excluded both diamond cards and a black king now.

So, the probability value is \(\begin{align} \dfrac{38}{52}\end{align}\)

which simplifies to \(\begin{align} \dfrac{19}{26} \end{align}\)

Probability of  getting '4' of diamonds = \(\begin{align} P(E)\:=\: \dfrac{1}{52}\end{align}\)

Probability of  getting a queen = \(\begin{align} P(E)\:=\: \dfrac{1}{13}\end{align}\)

Probability of  getting a black king = \(\begin{align} P(E)\:=\: \dfrac{1}{26}\end{align}\)

Probability of  getting a heart card = \(\begin{align} P(E)\:=\: \dfrac{1}{4}\end{align}\)

Probability of  getting a king or a queen = \(\begin{align} P(E)\:=\: \dfrac{2}{13}\end{align}\)

Probability of  getting a non-face card = \(\begin{align} P(E)\:=\: \dfrac{10}{13}\end{align}\)    

Probability of  getting neither a diamond or a black king \(\begin{align} P(E)\:=\: \dfrac{19}{26} \end{align}\)

 
Challenge your math skills
Challenging Questions
  1. The square of 10 is divided by 8. What is the probability of the remainder not to be an even number.
  2. In an exhibition, Edward and Eric throw a ring on a teddy bear. The probability of Edward to win the game is 1/8 and the probability of Eric to win the game is 1/6. Find out the probability so that only one of them is the winner. 

Interactive Questions

Here are a few activities for you to practice. Select/Type your answer and click the "Check Answer" button to see the result.
 

 
 
 
 
 
 

 


Let's Summarize

The mini-lesson targeted the fascinating concept of probability. The math journey around probability starts with what a student already knows, and goes on to creatively crafting a fresh concept in the young minds. Done in a way that not only it is relatable and easy to grasp, but also will stay with them forever. Here lies the magic with Cuemath.

About Cuemath

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Be it worksheets, online classes, doubt sessions, or any other form of relation, it’s the logical thinking and smart learning approach that we, at Cuemath, believe in.


Frequently Asked Questions(FAQs)

1. What is probability?

Probability is a branch of math which deals with finding out the likelihood of the occurrence of an event. The value of probability ranges between 0 and 1.

2. How to determine probability?

By dividing the number of outcomes of an event by the total number of possible outcomes or sample space we can determine probability.

3, What is conditional probability?

If there are two events A and B, conditional probability is a chance of occurrence of event B provided the event A has already occurred.

4. What is experimental probability?

Experimental probability is defined as the ratio of the total number of times an event has occurred to the total number of trials conducted.

5. What is a probability distribution?

It is a function in statistics that shows all the expected possible values for a set of data in a given range. 

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