Independent Events

Independent Events

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

Formula

P(A and B) = P(A)*P(B)

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

If a fair die and a coin is tossed together once, find the probability of getting a 3 and a head.

[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

If a fair die is tossed three times, find the probability of getting an even number three times.

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

Numbers from 1 to 10 are given. A number is randomly picked and replaced. Then a number is randomly picked again. Find the probability of picking the multiples of 3 twice.

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.