# 8. Independent and Dependent Events

If the occurrence or non-occurrence of *E*_{1} does not affect the probability of occurrence of *E*_{2}, then

P(E_{2}|E_{1}) =P(E_{2})

and *E*_{1} and *E*_{2} are said to be **independent events**.

Otherwise they are said to be **dependent** events.

[Recall from Conditional Probability that the notation *P*(*E*_{2} | *E*_{1}) means "the probability of the event *E*_{2} given that *E*_{1} has already occurred".]

## Two Events

Let's consider "*E*_{1} and *E*_{2}" as the event that "both *E*_{1} and *E*_{2} occur".

If *E*_{1} and *E*_{2} are **dependent events**, then:

P(E_{1}andE_{2}) =P(E_{1}) ×P(E_{2}|E_{1})

If *E*_{1} and *E*_{2} are **independent events**, then:

P(E_{1}andE_{2}) =P(E_{1}) ×P(E_{2})

## Three Events

For three **dependent **events *E*_{1}, *E*_{2}, *E*_{3}, we have

*P*(*E*_{1} and *E*_{2} and *E*_{3})

= *P*(*E*_{1}) × *P*(*E*_{2} | *E*_{1}) × *P*(*E*_{3} |* E*_{1} and *E*_{2})

For three **independent **events *E*_{1}, *E*_{2}, *E*_{3}, we have

*P*(*E*_{1} and *E*_{2} and *E*_{3}) = *P*(*E*_{1}) × *P*(*E*_{2}) × *P*(*E*_{3})

### Example 1

If the probability that person *A* will be alive in `20` years is `0.7` and the probability that person *B* will be alive in `20` years is `0.5`, what is the probability that they will both be alive in `20` years?

Answer

These are independent events, so

*P*(*E*_{1} and *E*_{2}) = *P*(*E*_{1}) × *P*(*E*_{2}) = 0.7 × 0.5 = 0.35

[Note, however, that if person *A* knows person *B*, then they will be **dependent** events, especially if *A* is married to *B*.]

Easy to understand math videos:

MathTutorDVD.com

### Example 2

A fair die is tossed twice. Find the probability of getting a `4` or `5` on the first toss and a `1`, `2`, or `3` in the second toss.

Answer

*P*(*E*_{1}) = *P*(4 or 5) = `2/6 = 1/3`

*P*(*E*_{2}) = *P*(1, 2 or 3) `= 3/6 = 1/2`

They are independent events, so

`P(E_1" and "E_2) ` `= P(E_1) × P(E_2) ` `= 1/3 × 1/2 ` `= 1/6`

Please support IntMath!

### Example 3

Two balls are drawn successively without replacement from a box which contains `4` white balls and `3` red balls. Find the probability that

(a) the first ball drawn is white and the second is red;

(b) both balls are red.

Answer

(a) The second event is dependent on the first.

P(E_{1}) =P(white) = `4/7`

There are 6 balls left and out of those 6, three of them are red. So the probability that the second one is red is given by:

P(E_{2}|E_{1}) =P(red) `= 3/6 = 1/2`

Dependent events, so

`P(E_1\ "and"\ E_2) = P(E_1) × P(E_2|E_1)` ` = 4/7 × 1/2 = 2/7`

(b) Also dependent events. Using similar reasoning, but realising there will be 2 red balls on the second draw, we have:

`P(R R) = 3/7 times 2/6 = 1/7`

Please support IntMath!

### Example 4

A bag contains `5` white marbles, `3` black marbles and `2` green marbles. In each draw, a marble is drawn from the bag and not replaced. In three draws, find the probability of obtaining white, black and green in that order.

Answer

We have 3 dependent events.

`P(W_1) times P(B_2` | `{:W_1) times ` `P(G_3 | {:B_2 " and " W_1)` `=5/10 times 3/9 times 2/8` `=1/24`

Please support IntMath!

### Search IntMath, blog and Forum

### Online Math Solver

This math solver can solve a wide range of math problems.

Go to: Online algebra solver

### Math Lessons on DVD

Math videos by MathTutorDVD.com

Easy to understand math lessons on DVD. See samples before you commit.

More info: Math videos

### The IntMath Newsletter

Sign up for the free **IntMath Newsletter**. Get math study tips, information, news and updates each fortnight. Join thousands of satisfied students, teachers and parents!