Events are dependent when the outcome of one event affect the outcome of a second event. Two events A and B in a probability space S are known to be dependent events if the occurrence of one affects the occurrence of the other. Suppose two cards are picked from a pack of playing cards without replacement, then the event that describes the type of the second card depends on the outcome of the first pick. The term **'without replacement'** present in a situation would mean the events involved are dependent.

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When two events are dependent, then the** outcome of one event affects the outcome of the other event. **In probability notation, the two events A and B are dependent if either P(A|B) $\neq$ P(A) or P(B|A) $\neq$ P(B). Two events A and B in a probability space are known to be dependent
events, if the occurrence of one affects the occurrence of the other.

Probability of two dependent events both occurring, multiply the probability of A and the probability of B after A occurs. If two events are dependent, we can generalize the probability formula for the compound event as follows:

**If two events A and B are dependent then**

P(A and B) = P(A) . P(B following A)

### Solved Example

**Question: **Two cards are picked at random from a pack of playing cards. What is the
probability of picking an ace first and then picking a face card,
picked card is not replaced.

** Solution: **

**Since the first card is not replaced we have a situation consisting of dependent events.**

**Step 1:**

Let** A **be the event of picking an ace in the first pick and** B **the event of the second card being a face card.

Total cards in an deck = 52

Face cards in a deck = 12

Number of ace = 4

**Step 2:**

P(A and B) = P(A) . P(B following A)

P(Ace and a face card) = P(Ace) . P(Face card after Ace)

**Case 1: **Probability of picked an ace from** 52 cards**

=> P(Ace) = $\frac{4}{52} = \frac{1}{13}$

**Case 2: **Probability of picked a face from **51 cards.**

There 12 face cards and they are to be picked from the 51 cards as the card picked is not replaced.

=> P(Face card after Ace) = $\frac{12}{51} = \frac{4}{17}$

**Step 3:**

P(Ace then a Face Card) = $\frac{1}{13}.\frac{4}{17} = \frac{4}{221}$.

If events are dependent, then occurrence of one event affects the probability of occurrence of other event. Often, more than two dependent events are combined to define a compound event. The probabilities are then found extending the method used for two dependent events.

### Solved Example

**Question: **A set of game tiles consists 5 each of dragon tiles, bird
tiles, flower tiles, dotted tiles, and line tiles. The game is to turn
up 5
tiles out of these 25 tiles which are placed with the pictures down. If
all the 5 tiles so turned are of same type, the person who turned them
will be awarded a cash
prize of 500 dollars. If the all 5 tiles turned up are of different
types, then the person would be awarded a cash prize of 200 dollars.
What are the probabilities of winning these two prices?

** Solution: **

The events are dependent here as the tiles turned up remain up.

Step 1:

To find the probability of turning 5 tiles of the same type.

We use the permutation counting techniques to determine the number of ways of turning 5 tiles of the same type and also the number of ways of turning 5 tiles out of 25 tiles turned down.

=> Number of ways of turning 5 tiles up from 25 turned down = P(25,5)

= $\frac{25!}{(25-5)!} $

= $\frac{25!}{20!}$

= 25.24.23.22.21

= 6,375,600.

And number of ways of turning 5 tiles of the same type one after the other = 5 x 4 x 3 x 2 x 1 = 120

=>**Probability of winning a cash prize of 500 dollars **= $\frac{120}{6375600}$ = 0.0000188.

**Step 2:**

To find the probability of turning all 5 different tiles.

There are 25 possibilities for turning the first tile up. Having turned the first one, there are 20 tiles which are of different type from the first tile turned up. Having turned two tiles of different types up, we have 15 tiles for the third turn. Applying this logic,

The number of ways of turning 5 tiles of all different types = 25 x 20 x 15 x 10 x 5 = 375,000.

Total number of ways turning 5 tiles from 25 turned down = 6,375,600

=>** Probability of winning a cash prize of 200 dollars **= $\frac{375,000}{6,375,600}$ = 0.0589.

Below you could see some examples of dependent events:

### Solved Examples

**Question 1: **A card is chosen at random from a deck of 52 playing cards. Without
replacing it, a second card is chosen. What is the probability that the
first card chosen is a king and the second card chosen is a queen?

** Solution: **

**Step 1:**

**Since 'without replacing cards', we have a situation consisting of dependent events.**

Total cards in an deck = 52

Number of king cards in a deck = 4

Number of queen cards = 4

**Step 2:**

P(A and B) = P(A) . P(B following A)

P(king and a queen) = P(king) . P(queen after king)

**Case 1: **Probability of picked a king card from** 52 cards.**

P(king on first pick) = $\frac{4}{52}$

**Case 2: **Probability of picked a queen card from** 51 cards.**

P(Queen on second pick after king) = $\frac{4}{51}$

P(king and queen) = $\frac{4}{52}$ * $\frac{4}{51}$ = $\frac{4}{663}$.

**Question 2: **On a math test, 6 out of 30 students got an grade A. If 3 students are
chosen at random without replacement, what is the probability that all
three got grade A on the test ?

** Solution: **

Number of students = 30

Students got grade A = 6

**Step 1:**

Probability of choosing 3 student out of 30, who got grade A.

**Case 1: **

Probability of choosing first student from 30 students.

P(first student) = $\frac{6}{30}$ = $\frac{1}{5}$

**Case 2: **

Probability of choosing second student from 29 students.

P(second student) = $\frac{5}{29}$

**Case 3: **

Probability of choosing third student from 28 students.

P(first student) = $\frac{4}{28}$ = $\frac{1}{7}$

**Step 2: **

P(3 students) = $\frac{1}{5}$ * $\frac{5}{29}$ * $\frac{1}{7}$

= $\frac{1}{203}$

Hence probability of choosing 3 student out of 30, who got grade A is $\frac{1}{203}$.

Probability of two dependent events both occurring, multiply the probability of A and the probability of B after A occurs. If two events are dependent, we can generalize the probability formula for the compound event as follows:

P(A and B) = P(A) . P(B following A)

Let

Total cards in an deck = 52

Face cards in a deck = 12

Number of ace = 4

P(A and B) = P(A) . P(B following A)

P(Ace and a face card) = P(Ace) . P(Face card after Ace)

=> P(Ace) = $\frac{4}{52} = \frac{1}{13}$

=> P(Face card after Ace) = $\frac{12}{51} = \frac{4}{17}$

P(Ace then a Face Card) = $\frac{1}{13}.\frac{4}{17} = \frac{4}{221}$.

If events are dependent, then occurrence of one event affects the probability of occurrence of other event. Often, more than two dependent events are combined to define a compound event. The probabilities are then found extending the method used for two dependent events.

The events are dependent here as the tiles turned up remain up.

Step 1:

To find the probability of turning 5 tiles of the same type.

We use the permutation counting techniques to determine the number of ways of turning 5 tiles of the same type and also the number of ways of turning 5 tiles out of 25 tiles turned down.

=> Number of ways of turning 5 tiles up from 25 turned down = P(25,5)

= $\frac{25!}{(25-5)!} $

= $\frac{25!}{20!}$

= 25.24.23.22.21

= 6,375,600.

And number of ways of turning 5 tiles of the same type one after the other = 5 x 4 x 3 x 2 x 1 = 120

=>

To find the probability of turning all 5 different tiles.

There are 25 possibilities for turning the first tile up. Having turned the first one, there are 20 tiles which are of different type from the first tile turned up. Having turned two tiles of different types up, we have 15 tiles for the third turn. Applying this logic,

The number of ways of turning 5 tiles of all different types = 25 x 20 x 15 x 10 x 5 = 375,000.

Total number of ways turning 5 tiles from 25 turned down = 6,375,600

=>

Total cards in an deck = 52

Number of king cards in a deck = 4

Number of queen cards = 4

P(A and B) = P(A) . P(B following A)

P(king and a queen) = P(king) . P(queen after king)

P(king and queen) = $\frac{4}{52}$ * $\frac{4}{51}$ = $\frac{4}{663}$.

Number of students = 30

Students got grade A = 6

Probability of choosing first student from 30 students.

Probability of choosing second student from 29 students.

Probability of choosing third student from 28 students.

P(3 students) = $\frac{1}{5}$ * $\frac{5}{29}$ * $\frac{1}{7}$

= $\frac{1}{203}$

Hence probability of choosing 3 student out of 30, who got grade A is $\frac{1}{203}$.

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