# Mutually Non-Exclusive Events

Definition

of Mutually Non-Exclusive Events:

Two events A and B are said to be mutually non exclusive events if both the

events A and B have atleast one common outcome between them.

The events A and B cannot prevent the occurrence of one another so from

here we can say that the events A and B have something common in them.

**For example,** in the case of rolling a die the event of getting an ‘odd-face’ and the event of getting ‘less than 4’ are not mutually exclusive and they are also known as compatible event.

The event of getting an ‘odd-face’ and the event of getting ‘less than 4’ occur when we get either 1 or 3.

Let ‘X’ is denoted as event of getting an ‘odd-face’ and

‘Y’ is denoted as event of getting ‘less than 4’

The events of getting an odd number (X) = {**1**, **3**, 5}

The events of getting less than 4 (Y) = {**1**, 2, **3**}

Between

the events X and Y the common outcomes are 1 and 3

**Therefore, the events X and Y are compatible events/mutually
non-exclusive.**

Addition Theorem Based on Mutually Non-Exclusive Events:

If X and Y are two mutually Non- Exclusive Events, then the probability of ‘X union Y’ is the difference between the sum of the probability of X and the probability of Y and the probability of ‘X intersection Y’ and represented as,

**P(X ∪ Y) = P(X) + P(Y) – P(X ∩ Y)**

**Proof:** The events X – XY, XY and Y – XY are pair-wise mutually exclusive events then,

X = (X – XY) + XY,

Y = XY + (Y – XY)

Now, P(X) = P(X – XY) + P(XY)

or, P(X – XY) = P(X) – P (XY)

Similarly, P(Y – XY) = P(Y) – P(XY)

Again, P(X + Y) = P(X – XY) + P(XY) + P(Y – XY)

⇒ P(X + Y) = P(X) – P(XY) + P(XY) + P(Y) – P(XY)

⇒ P(X + Y) = P(X) + P(Y) – P(XY)

⇒ P(X + Y) = P(X) + P(Y) – P(X) P(Y)

**Therefore, P(X ∪ Y) = P(X) + P(Y) – P(X ∩ Y)**

Worked-out problems on probability of Mutually Non-Exclusive Events:

**1.** What is the probability of getting a diamond or a queen from a well-shuffled deck of 52 cards?

**Solution:**

Let X be the event of ‘getting a diamond’ and,

Y be the event of ‘getting a queen’

We know that, in a well-shuffled deck of 52 cards there are 13 diamonds and 4 queens.

Therefore, probability of getting a diamond from well-shuffled deck of 52 cards = P(X) = 13/52 = 1/4

The probability of getting a queen from well-shuffled deck of 52 cards = P(Y) = 4/52 = 1/13

Similarly, the probability of getting a diamond queen from well-shuffled deck of 52 cards = P(X ∩ Y) = 1/52

According to the definition of mutually non-exclusive we know that, drawing of a well-shuffled deck of 52 cards ‘getting a diamond’ and ‘getting a queen’ are known as mutually non-exclusive events.

We have to find out Probability of X union Y.

So according to the addition theorem for mutually non- exclusive events, we get;

**P(X ∪ Y) = P(X) + P(Y) – P(X ∩ Y)**

Therefore, P(X U Y) |
= 1/4 + 1/13 – 1/52
= (13 + 4 – 1)/52 = 16/52 = 4/13 |

Hence, probability of getting a diamond or a queen from a well-shuffled deck of 52 cards = 4/13

**2.** A

lottery box contains 50 lottery tickets numbered 1 to 50. If a lottery ticket

is drawn at random, what is the probability that the number drawn is a multiple

of 3 or 5?

Solution:

Let X be the event of

‘getting a multiple of 3’ and,

Y be the event of

‘getting a multiple of 5’

The events of getting a multiple of 3 (X) = {3,6,9,12,**15**,18,21,24,27,**30**,

33,36,39,42,**45**,48}

Total

number of multiple of 3 =

16

P(X) = 16/50 = 8/25

The events

of getting a multiple of 5 (Y) = {5, 10, **15**, 20, 25, **30**, 35, 40, **45**, 50}

Total

number of multiple of 3 =

16

P(X) = 10/50 = 1/5

Between

the events X and Y the favorable outcomes are 15, 30 and 45.

Total

number of common multiple

of both the number 3 and 5 = 3

The probability

of getting a ‘multiple of

3’ and a ‘multiple

of 5’ from the numbered 1 to 50 = P(X ∩ Y) = 3/50

Therefore, X and Y are non mutually exclusive events.

We have to find out Probability

of X union Y.

So according to the

addition theorem for mutually non- exclusive events, we get;

**P(X ∪ Y) = P(X) + P(Y) – P(X ∩ Y)**

Therefore, P(X U Y) |
= 8/25 + 1/5 – 3/50
= (16 + 10 = 23/50 |

**Hence, probability of
**

**getting**

**multiple of 3 or 5 = 23/50**