Conditional Probability

Introduction

Conditional probability is defined as the likelihood of an event or outcome occurring, based on the occurrence of a previous event or outcome. In this tutorial, we will learn about Probability, Dependent and Independent Events, Conditional Probability and Conditional Probability theorem. We will also learn about Multiplication Rule in Probability and Total Probability and Bayes’ theorem.

The probability of an event occurring given that another event has already occurred is known as conditional probability. For Conditional probability the events has to be dependent. Dependent events are the events where the occurrence of one event causes change in the probability of occurrence of another event.

Conditional Probability

The probability of an event occurring given that another event has already occurred is known as the conditional probability of the first event given the second event. It is represented by ‘|’.

E.g. Let A and B be the event of the sum of the throw of a dice two times is 12 and in the first throw getting 6 respectively.

Now, we know that the Probability of 6 occurring on a dice throw is 1 of 6.

i.e, $\mathrm{P(B)\:=\:\frac{1}{6}}$

And the probability of the sum of 12 on two throws is 1 of 36.

i.e, $\mathrm{P(A)\:=\:\frac{1}{36}}$

But, if the first die throw already has a 6, then for the sum of 12 the second die only needs 6 on it, i.e. the probability of A given B is 1 of 6

i.e, $\mathrm{P(A/B)\:=\:\frac{1}{6}}$

The conditional probability is calculated by the following formula,

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

In this example the events A and B are dependent on each other. These types of events are known as Dependent Events.

Dependent Events are defined as the events where the occurrence of one has an effect on the probability of occurrence of the other event.

Conditional Probability Theorems and Properties.

Property 1 − Let E be an event of a sample space S of an experiment, then we have −

$\mathrm{P(S|E)\:=\:P(E|E)\:=\:1}$$

Proof/Explanation − The probability of E given E has already occurred is trivially 1.

Now for the other part, the E is an event in the sample space S, in set notations this is represented by 𝐸 ⊆ 𝑆 (E is a subset of S). Now if E occurs that means a part of the sample space S has occurred that means S as an event has occurred. Therefore, the probability of S given E is also 1.

Property 2 − If A and B are any two events of a sample space S and F is an event of S such that P(F) ≠ 0, then;

$\mathrm{P((A\cup\:B)|F)\:=\:P(A|F)\:+\:P(B|F)\:-\:P((A\cap\:B)|F)}$

Proof/Explanation −

We know,

$$\mathrm{P(A\cup\:B)\:=\:P(A)\:+\:P(B)\:-\:P(A\cap\:B)}$$

Also,$\mathrm{P(A|F)\:=\:\frac{P(A\cap\:F)}{P(F)}}$

By distributive property, we can perform the following operation on both sides.

$\mathrm{P(A\cup\:B)\:P(A)\:+\:P(B)\:-\:P(A\cap\:B)\:\:\:\:\:\:\:\:\:\:\:\:...\:\:\:\:\:\:\:\:\:\cap\:F\:on\:both\:sides}$

$\mathrm{P((A\cup\:B)\cup\:F)\:=\:P(A\cap\:F)\:+\:P(B\cap\:F)\:-\:P((A\cap\:B)\cap\:F)}$

Also, since $\mathrm{P(F)\:\neq\:0}$, divide both sides of the equation by P(F)

$\mathrm{P((A\cup\:B)\cap\:F)/P(F)\:=\:P(A\cap\:F)/P(F)\:+\:P(B\cap\:F)/P(F)\:-\:P((A\cap\:B)\cap\:F)/P(F)}$

$\mathrm{\Longrightarrow\:P((A\:\cup\:B)|F)\:=\:P(A|F)\:+\:P(B|F)\:-\:P((A\cap\:B)|F)}$

Property 3 − $\mathrm{P(A'|B)\:=\:1\:-\:P(A|B)}$

Proof/Explanation −

We know that

$\mathrm{P(A')\:=\:1\:-\:P(A)}$

We also know that

$\mathrm{P(S)\:=\:1}$

Therefore we can rewrite the equation as

$\mathrm{P(A')\:=\:P(S)\:-P(A)}$

Similar to the last property, we can use the distributive property of intersection,

$\mathrm{P(A'\cap\:B)\:=\:P(S\cap\:B)\:-\:P(A\cap\:B)}$

Dividing by P(B),

$\mathrm{P(A'\cap\:B)/P(B)\:=\:P(s\cap\:B)/P(B)\:-\:P(A\cap\:B)/P(B)}$

$\mathrm{P(A'|B)\:=\:P(S|B)\:-\:P(A|B)}$

Using Property 1, $\mathrm{P(S|B)\:=\:1}$

$\mathrm{P(A'|B)\:=\:1\:-\:P(A|B)}$

Multiplication Rule in Probability

The multiplication rule in probability has two forms, one for dependent events and another for independent events.

Multiplication Rule for dependent events:

The Conditional probability of A given B is given by,

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

Rearranging the terms we have,

$$\mathrm{P(A\cap\:B)\:P(B)\times\:P(A|B)}$$

Similarly

$$\mathrm{P(A\cap\:B)\:=\:P(A)\times\:P(B|A)}$$

Multiplication Rule for independent events:

The Conditional probability of A given B is given by,

$$\mathrm{P(A|B)\:=\:P(B)\:P(B)\times\:P(A|B)}$$

But we know, that independent events don’t affect each other’s occurrence

$$\mathrm{\Longrightarrow\:P(A|B)\:=\:P(A)\:and\:P(B|A)\:=\:P(B)}$$

Substituting P(A) for P(A|B)

$$\mathrm{\Longrightarrow\:P(A\cap\:B)\:=\:P(B)\times\:P(A)}$$

Total Probability and Bayes Theorem

Total Probability Theorem

he total Probability theorem states that, If A is an event in a sample space S, which can be partitioned into n Independent events, say 𝐸1, 𝐸2, … , 𝐸𝑛, all mutually dependent with A, then the probability of A is given by

$$\mathrm{P(A)\:=\:P(A|E_{1}).P(E_{1})\:+\:P(A|E_{2}).P(E_{2})\:+\:.......\:+\:P(A|E_{n}).P(E_{n})}$$

$$\mathrm{\Longrightarrow\:P(A)\:=\:\displaystyle\sum\limits_{i=1}^n P(A|E_{i}).P(E_{i})}$$

Proof −

The probability of Event A in terms of its intersection with the events $\mathrm{E_{i},i\:=\:1,2,3,...,n}$ is given by

$$\mathrm{P(A)\:=\:P(A\cap\:E_{1})\:+\:P(A\cap\:E_{2})\:+\:.....\:+\:P(A\cap\:E_{i})}$$

Using the Multiplication Rule of Probability for dependent events, we have,

$$\mathrm{P(A\cap\:E_{i})\:=\:P(A|E_{i}).P(E_{i}),\forall\:i\:=\:1,2,3,.......,n}$$

Substituting this in the previous equation we have,

$$\mathrm{P(A)\:=\:P(A|E_{1}).P(E_{1})\:+\:P(A|E_{2}).P(E_{2})\:+\:.......\:+\:P(A|E_{n}).P(E_{n})}$$

$$\mathrm{\Longrightarrow\:P(A)\:\displaystyle\sum\limits_{i=1}^n P(A|E_{i}).P(E_{i})}$$

Bayes Theorem

The Bayes theorem relates the probability of the first event given second to the probability of the second event given first. If A is an event in a sample space S, which can be partitioned into n Independent events, say 𝐸1, 𝐸2, … , 𝐸𝑛, all mutually dependent with A, then the probability of any 𝐸𝑖 given A, given the probability of A given $\mathrm{E_{i},\forall\:i\:=\:1,2,3,.....,n.}$ is given by,

$$\mathrm{P(E_{i}|A)\:=\:\frac{P(A|E_{i}).P(E_{i})}{P(A|E_{1}).P(E_{1})\:+\:P(A|E_{2}).P(E_{2})\:+\:.......\:+\:P(A|E_{n}).P(E_{n})}}$$

Proof −

We know from the conditional probability formula,

$$\mathrm{P(E_{i}|A)=\:\frac{P(E_{i}\cap\:A)}{P(A)}\:=\:\frac{P(A\cap\:E_{i})}{P(A)}}$$

Also by the Multiplication Rule for dependent Events,

$$\mathrm{P(A\cap\:E_{i})\:=\:P(A|E_{i}).P(E_{i})}$$

And By total probability theorem

$$\mathrm{P(A)\:=\:P(A|E_{i}).P(E_{i})\:+\:P(A|E_{2}).P(E_{2})\:+\:.......\:+\:P(A|E_{n}).P(E_{n})}$$

Substituting these in the equation we have,

$$\mathrm{P(E_{i}|A)\:=\:\frac{P(A|E_{i}).P(E_{i})}{P(A|E_{1}).P(E_{1})\:+\:P(A|E_{2}).P(E_{2})\:+\:.......\:+\:P(A|E_{n}).P(E_{n})}}$$

Solved Examples

1) Let there be two identical bags B1 and B2, both bags contain some blue and some red balls (B1 (6Blue, 4 Red), B2 (5Blue, 5Red)). If a bag is chosen at random and a ball is taken out of the bag, and it turns out to be blue, then what is the probability that the bag chosen was B2?

Answer − Let the B1 and B2 be the events that bags B1 and B2 are chosen respectively

And let A be the event that a ball drawn from the chosen bag is Blue

Then,$\mathrm{P(B_{1})\:=\:P(B_{2})\:=\:0.5}$

$\mathrm{P(A|B_{1})\:=\:0.6,P(A|B_{2})\:=\:0.5}$

Then by bayes theorem

$$\mathrm{P(B_{2}|A)\:=\:\frac{P(A|B_{2}).P(B_{2})}{P(A|B_{1}).P(B_{1})\:+\:P(A|B_{2}).P(B_{2})}\:=\:\frac{0.5\times\:0.5}{(0.6\times\:0.5)\:+\:(0.5\times\:0.5)}}$$

$$\mathrm{P(B_{2}|A)\:=\:\frac{0.25}{0.3\:0.25}\:=\:\frac{0.25}{0.55}\:=\:\frac{22}{55}}$$

$$\mathrm{P(B_{2}|A)\:=\:\frac{5}{11}\:=\:0.4545....}$$

Conclusion

In this tutorial, we will learn about Probability, Dependent and Independent Events, Conditional Probability and Conditional Probability theorem. We will also learn about Multiplication Rule in Probability and Total Probability and Bayes’ theorem.

FAQs

1. What do you mean by probability?

Probability is the mathematical representation of the chance of a well defined event’s occurrence.

2. What is a Sample Space?

Sample space is defined as the collection of all possible outcomes of an experiment

3. Define conditional probability?

The probability of an event occurring, given that another event has already occurred, is known as conditional probability

4. What is the multiplication rule in probability?

It states that the probability of two Independent events occurring simultaneously is the product of the probability of those events occurring individually.