# Independent Events

How to find the probability of independent events: formula, examples, and their solutions.

## Formula

Independent events are the events

that don't affect each other.

If *A* and *B* are independent events,

then P(*A* and *B*) can be found

by multiplying P(*A*) and P(*B*):

P(*A* and *B*) = P(*A*)⋅P(*B*).

P(*A* and *B*): Probability of both *A* and *B* happening

P(*A*): Probability of *A* happening

P(*B*): Probability of *B* happening

Probability of (*A* and *B*, Intersection)

## Example 1

[Getting a 4, *A*] and [getting a head, *B*]

don't affect each other.

So these two events are independent events.

So find P(*A*) and P(*B*) separately,

then find P(*A* and *B*)

by multiplying the probabilities.

First, see the fair die.

A die has 6 sides: from 1 to 6.

So there are 6 ways to get a number.

So n(*S*) = 6.

Set the event *A* as

getting a 4.

So write {4}.

So n(*A*) = 1.

n(*S*) = 6

n(*A*) = 1

So P(*A*) = 1/6.

Probability

Next, see the coin.

A coin has 2 sides: Head or Tail.

So n(*S*) = 2.

Set the event *B* as

getting a head.

So write {H}.

So n(*B*) = 1.

n(*S*) = 2

n(*B*) = 1

So P(*B*) = 1/2.

P(*A*) = 1/6

P(*B*) = 1/2

So P(*A* and *B*) = (1/6)⋅(1/2).

So 1/12 is the answer.

## Example 2

Each toss don't affect each other.

So the three tosses are independent events.

So find P(*A*), P(*B*), and P(*C*) separately,

then find P(*A* and *B* and *C*)

by multiplying the probabilities.

First, see the first toss.

A die has 6 sides: from 1 to 6.

So there are 6 ways to get a number.

So n(*S*) = 6.

Set the event *A* as

getting an even number at the first toss.

The even numbers from 1 to 6 are

{2, 4, 6}.

So n(*A*) = 3.

n(*S*) = 6

n(*A*) = 3

So P(*A*) = 3/6

= 1/2.

Probability

Next, see the second toss.

Set the event *B* as

getting an even number at the second toss.

Basically, *A* and *B* are the same event:

getting an even number from a fair die.

So n(*S*) is the same: 6.

And n(*A*) = n(*B*) = 3.

So P(*B*) = P(*A*) = 1/2.

This is the same for the third toss.

n(*S*) is the same: 6.

And n(*A*) = n(*B*) = n(*C*) = 3.

So P(*C*) = 1/2.

P(*A*) = P(*B*) = P(*C*) = 1/2

So P(*A* and *B* and *C*) = (1/2)⋅(1/2)⋅(1/2).

So 1/8 is the answer.

## Example 3

After the first pick,

the picked number is replaced.

So the first pick and the second pick

don't affect each other.

So these two events are independent events.

So find P(*A*) and P(*B*) separately,

then find P(*A* and *B*)

by multiplying the probabilities.

First see the first pick.

Numbers from 1 to 10 are given.

So there are 10 numbers

that can be picked.

So n(*S*) = 10.

Set the event *A* as

picking the multiples of 3 at the first pick.

The multiples of 3 from 1 to 10 are

{3, 6, 9}.

So n(*A*) = 3.

n(*S*) = 10

n(*A*) = 3

So P(*A*) = 3/10.

Probability

Next, see the second pick.

Set the event *B* as

picking the multiples of 3 from 1 to 10.

Then you can say that*A* and *B* are the same event:

getting a multiple of 3 from 1 to 10.

So n(*S*) is the same: 10.

And n(*A*) = n(*B*) = 3.

So P(*B*) = P(*A*) = 3/10.

P(*A*) = P(*B*) = 3/10

So P(*A* and *B*) = (3/10)⋅(3/10).

So 9/100 is the answer.