# 8. Independent and Dependent Events

If the occurrence or non-occurrence of E1 does not affect the probability of occurrence of E2, then

P(E2 | E1) = P(E2)

and E1 and E2 are said to be independent events.

Otherwise they are said to be dependent events.

[Recall from Conditional Probability that the notation P(E2 | E1) means "the probability of the event E2 given that E1 has already occurred".]

## Two Events

Let's consider "E1 and E2" as the event that "both E1 and E2 occur".

If E1 and E2 are dependent events, then:

P(E1 and E2) = P(E1) × P(E2 | E1)

If E1 and E2 are independent events, then:

P(E1 and E2) = P(E1) × P(E2)

## Three Events

For three dependent events E1, E2, E3, we have

P(E1 and E2 and E3)

= P(E1) × P(E2 | E1) × P(E3 | E1 and E2)

For three independent events E1, E2, E3, we have

P(E1 and E2 and E3) = P(E1) × P(E2) × P(E3)

Continues below

### 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?

These are independent events, so

P(E1 and E2) = P(E1) × P(E2) = 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.]

### 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.

P(E1) = P(4 or 5) = 2/6 = 1/3

P(E2) = 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

### 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.

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

P(E1) = 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(E2 | E1) = 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

### 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.

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