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Reference

Independent and Dependent Events

Concept

Independent Events

Two events $A$ and $B$ are independent events if the occurrence of either of these events does not affect the occurrence of the other. It is also said that they are independent if and only if the probability that both events occur is equal to the product of the individual probabilities.

P (A \cap B) = P (A) \cdot P (B)

Why

For example, consider drawing two marbles from a bowl, one at a time.

The probability of first picking a green marble can be calculated by dividing the favorable outcomes by the possible outcomes. Let

G,

B,

and

O

be the events of drawing green, blue, and orange marbles, respectively. There is

1

green marble and

3

marbles in total.

P (G) = \frac{1}{3}

Suppose that the first marble is replaced before the second draw. Therefore, after the replacement there is

1

orange marble, and

3

marbles in total.

P (O) = \frac{1}{3}

Note that there are

9

possible outcomes for drawing two marbles one at a time. Only

1

of these options corresponds to an event of drawing a green marble and then an orange marble.

G G G B G O B B B G B O O O O G O B

Therefore, the combined probability of picking a green marble first and an orange marble second is

\frac{1}{9} .

Since the probability that both events occur is equal to the product of the individual probabilities, these events can be considered as independent events.

\frac{1}{3} \cdot \frac{1}{3} = \frac{1}{9} ⇓ P (G) \cdot P (O) = P (G \cap O)

Concept

Dependent Events

Two events $A$ and $B$ are considered dependent events if the occurrence of either of these events affects the occurrence of the other. If the events are dependent, the probability that both events occur is equal to the product of the probability of the first event occurring and the probability of the second event occurring after the first event.

P (A \cap B) = P (A) \cdot P (B ∣ A)

Why

For example, consider drawing two marbles from a bowl, one at a time.

The probability of first picking a green marble can be calculated by dividing the favorable outcomes by the possible outcomes. Let

G,

B,

and

O

be the events of drawing green, blue, and orange marbles, respectively. There is

1

green marble and

3

marbles in total.

P (G) = \frac{1}{3}

Suppose that after the green marble is picked, it is not replaced in the bowl.

This affects the probability of picking an orange marble on the second draw. Now there is still

1

orange marble, but instead of

3,

there are

2

marbles in total.

P (O ∣ G) = \frac{1}{2}

Using this information, the sample space of the described situation can be found.

G B G O B G B O O G O B

Out of

6,

there is only

1

outcome that corresponds to first drawing a green marble and then an orange marble. Therefore, the probability of picking a green and then an orange marble is

\frac{1}{6} .

\frac{1}{3} \cdot \frac{1}{2} = \frac{1}{6} ⇓ P (G) \cdot P (O ∣ G) = P (G \cap O)

Because the occurrence of the first event affects the occurrence of the second, these events can be concluded to be dependent.

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