 # Dependent Events in Probability

Dependent events in probability means events whose occurrence  of one affect the probability of occurrence of the other. For example suppose a bag has 3 red and 6 green balls. Two balls are drawn from the bag one after the other. Let A be event of drawing red ball in the first draw and B be the event of drawing green ball in the second draw. If the ball drawn in the first draw is not replaced back in the bag, then A and B are dependent events because P(B) is decreased or increased according to the first draw results as a red or green ball.

Probability theory is an important concept that is quite familiar to the students who study mathematics in higher classes. For example Weather forecast of some area says that there is a twenty percent probability that it will rain tomorrow.

The probability is a chance of some event to happen. The term “event” actually means one or even more outcomes. The event is described as the outcome which is able to occur. Total events are defined as all the outcomes which may possibly occur relevant to the experiment asked in the question. Also, the events of interest are known as favorable events. More elaborately, in the branch of probability, an event is defined to be the set of all the possible outcomes for an experiment.

For example:

i) Obtaining a head in a toss of a coin may be called an event.

ii) Getting a 6 on a roll of a die is said to be an event.

(iii) Drawing an ace from the deck of cards is also an event.

(iv) Getting a sum of 9 on the roll of a pair of dice is an event.

An event whose chances of happening is 100 % is called a sure event. The probability of such an event is 1. In a sure event, one is likely to get the desired output in the whole sample experiment. On the other hand, when there are no chances of an event happening, the probability of such an event is likely to be zero. This is said to be an impossible event.

On the basis of quality events, these are classified into three types which are as follows:

A) Independent Events

B) Dependent Events

C) Mutually-Exclusive Events

## Dependent Events Definition

Dependent events are those which depend upon what happened before. These events are affected by the outcomes that had already occurred previously. i.e. Two or more events that depend on one another are known as dependent events. If one event is by chance changed, then another is likely to differ.

 Important Result: When two events, A and B are dependent, the probability of occurrence of A and B is: P(A and B) = P(A) · P(B|A)

For instance:

i) Let’s say three cards are to be drawn from a pack of cards. Then the probability of getting an ace is highest when the first card is drawn, while the probability of getting an ace would be less when the second card is drawn. In the draw of the third card, this probability would be dependent upon the outcomes of the previous two cards. We can say that after drawing one card, there will be fewer cards available in the deck, therefore the probabilities tend to change.

ii) Suppose that we want to have a queen. With the first draw of a card, the chances of getting a queen are 4 out of 52 cards. If we get a queen in the first draw, then the probability of getting queen in the second draw will be 3 out of 51 cards. Thus, these are said to be the dependent events, since the probability of the second event depends on the outcome of the first draw.

### Related articles

Probability of independent events

Bayes theorem of probability

Probability problems

## Solved Examples

Example 1: Shareen has to select two students from a class of 23 girls and 25 boys. What is the probability that both students chosen are boys?

Solution: Total number of students = 23 + 25 = 48

Probability of choosing the first boy, say Boy 1 = 25/48

Probability of choosing second boy, say Boy 2 = 24/47

Now,

P(Boy 1 and Boy 2) = P(Boy 1) and P(Boy 2|Boy 1)

= 25/48 . 24/47

= 600/2256

Example 2: In a survey found that 10 out of 13 people walk to the office. 3 persons are selected randomly. What is the probability that all three walks to the office?

Solution:

The probability that all three walk to the office = 3/13 * 2/12 * 1/11 = 6/1716

Example 3: A bag contains 6 red,  5 blue, and 4 yellow balls. 2 balls are drawn, but the first ball is drawn without replacement. Find the following.

a] P (red, then blue)

b] P (blue, then blue)

Solution:

a] There are six red balls and a total of fifteen balls.

P (red) = 6 / 15

The probability of the second draw affected the first.

Number of blue balls = 5

Total number of balls left = 14

P (drawing blue after red) = 5 / 14

P(drawing red, then blue) = P (drawing red) * P (blue after red) = $\frac{6}{15}*\frac{5}{14}=\frac{1}{7}$

1/ 7 will be the probability value of drawing a red ball followed by a blue ball.

b] Number of blue balls = 5

Total number of balls left = 15

The probability of drawing a blue ball = 5 / 15

The probability of the second draw affected the first.

Now there are 4 blue balls left and a total of 14 balls left.

P (drawing a blue ball after a blue ball) = 4 / 14

P (blue, then blue) = P (drawing blue ball) * (drawing a blue ball after a blue ball) = $\frac{5}{15}*\frac{4}{14}=\frac{2}{21}$

Hence, the probability of drawing a red ball followed by a blue ball is 2 / 21.

Example 4: In a pack of 52 cards, a card is drawn at random without replacement. Find the probability of drawing a queen followed by a jack.

Solution:

P (drawing a queen in the first place) = 4 / 52

P (drawing jack in the second place given that queen is in the first place) = 4 / 51

P (drawing a queen followed by a jack) = $\frac{4}{52}*\frac{4}{51}=\frac{16}{2652}=\frac{4}{663}$

Example 5: A wallet contains 4 bills of 5 dollars, 5 bills of 10 dollars and 3 bills of 20 dollars. 2 bills are chosen randomly without replacement. Find the P (drawing a 5 dollar bill followed by a 5 dollar bill).

Solution:

Number of 5 dollar bills = 4

Total number of bills = 12

P (drawing a 5 dollar bill) = 4 / 12

The probability of the second draw affected the first.

Number of 5 dollar bills left = 3

A total of 11 bills are left.

P (drawing a 5 dollar bill after a 5 dollar bill) = 3 / 11

P (drawing a 5 dollar bill followed by a 5 dollar bill) = P (drawing a 5 dollar bill) * P (drawing a 5 dollar bill after a 5 dollar bill) = $\frac{4}{12}*\frac{3}{11}=\frac{1}{11}$

Hence the P (drawing a 5 dollar bill followed by a 5 dollar bill) = 1 / 11

Example 6 : A bag contains 4 red, 3 pink and 6 green balls. Two balls are drawn, but the first ball drawn is not replaced.

a) Find P(red, then pink)

b) Find P(pink, then pink)

Solution:

a) number of red balls = 4

Total number of balls = 13

P(red) = 4/13

The result of the first draw will affect the probability of the second draw.

There are 3 pink balls.

Number of balls left = 12

P(pink after red) = 3/12 = ¼

P(red, then pink) = P(red) · P(pink after red) = (4/13)(¼) = 1/13

Hence  probability of drawing a red ball and then a pink ball is 1/13.

b)number of pink balls = 3

Total number of balls = 13

P(pink) = 3/13

The result of the first draw will affect the probability of the second draw.

Number of pink balls left = 2

Total number of balls left = 12

There are a total of 12 balls left.

P(pink after pink) = 2/12 = 1/6

P(pink, then pink) = P(pink) · P(pink after pink) = (3/13)(⅙) = 3/78

Hence  probability of drawing a pink ball and then a pink ball is = 3/78