# Independent and Dependent Events

The concept of independent and dependent events comes into play when we are working on conditional probability. A compound or joint events is the key concept to focus in conditional probability formula. Drawing a card repeatedly from a deck of 52 cards with or without replacement is a classic example to explain these concepts.

## Independent and Dependent Events

### Independent Events

Two independent events as disjoint sets; Ω denotes Sample Space.

An event that does not affect the occurrence of another subsequent event in a random experiment is an independent event.

• Ex. Tossing a coin.
• Here, Sample Space S =  {H, T}  and both H and T are independent events.
• Ex. Rolling a dice.
• Sample Space S = {1, 2, 3, 4, 5, 6}, all of these events are independent too.

Both of the above examples are simple events (events with single outcomes). Even compound events  (two events occurring at the same time) can be independent events.

• Ex. Tossing a coin and rolling a die.
• Sample space S = {(1,H), (2, H), (3, H), (4, H), (5, H), (6, H), (1, T), (2, T), (3, T), (4, T) (5, T) (6, T)}.
• These events are independent because only one can occur at a time.

### Dependent Events

When the occurrence of one event affects the occurrence of another subsequent event, the two events are dependent events. The concept of dependent events gives rise to the concept of conditional probability.

## Conditional Probability Formula

It is the probability of an event given that another event has already occurred. The event that occurred earlier may affect the probability of the present event under consideration. Also, the event happening later can occur only after the given former event has taken place.

### How to denote?

If the probability of events A and B are P(A) and P(B) respectively then the conditional probability of B such that A has already occurred is P(A/B).

### How to calculate?

P(A/B) = P(A ∩ B)/P(A) or P(B ∩ A)/P(A)  given, P(A)>0

P(A)<0 means A is an impossible event. In P(A ∩ B) the intersection denotes a compound probability.

### Illustration

• From a deck of 52 cards, a card is drawn randomly. We want to find the probability of getting a red card
• Let us denote this event as R1 and the probability as P(R1)
• Now if we again draw a card from the same deck what is the probability of getting a red card given that the card drawn earlier came out to be black and was not replaced into the deck
• The probability of getting a red this time is denoted as P(R2)
• conditional probability is P(R2/R1) = P(R1 ∩ R2) / P(R1)
• P(R1 ∩ R2) = P(R1) × P(R2)
• In case, replacement is done then P(R2/R1) = P(R2)

from the above information we can get the following relation:

• P(A ∩ B) = P(B/A) × P(A) ;  P(A)>0

## Solved Example on Conditional Probability Formula

Question. Out of 10 customers in a box 4 are purchased red bulbs, 3 purchased green bulbs and 2 purchased both green and red. If a customer at random bought a red bulb what is the probability that he/she also bought the green bulb?

Solution. The probability of consumers purchasing red from the question is P(R) = 4/10.

The probability of consumers purchasing both red and green from the question is P(R ∩ G) = 2/10.

So, from the given formula, the probability that a consumer buying red bulb is also buying the green bulb is

P(G/R) = P(G ∩ R)/P(R)

= 0.2/0.4

= 1/2 or 0.5

This concludes our discussion on the topic of conditional probability formula.

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