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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.


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`

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