# Mutually Exclusive Events

Definition of Mutually Exclusive Events:

If two events are such that

they cannot occur simultaneously for any random experiment are said to be mutually

exclusive events.

If X and Y are two mutually exclusive events, then X ∩ Y = ∅

**For
example,** events in rolling of a

die are “even face” and “odd face” which are known as mutually exclusive events.

But” odd-face” and

“multiple of 3” are not mutually exclusive, because when “face-3” occurs both

the events “odd face” and “multiply of 3” are said to be occurred

simultaneously.

We see that two

simple-events are always mutually exclusive while two compound events may or

may not mutually exclusive.

Addition Theorem Based on Mutually Exclusive Events:

If **X**

and **Y** are two mutually exclusive

events, then the probability of ‘X union Y’ is the sum of the probability of X and

the probability of Y and represented as,

**P(X U Y) = P(X) + P(Y)**

**Proof:** Let E be a random experiment and N(X) be

the number of frequency of the event X in E. Since **X** and **Y** are two mutually exclusive events

then;

N(X **U** Y) = N(X) +

N(Y)

or, N(X **U** Y)/N =

N(X)/N + N(Y)/N; Dividing both the sides by N.

Now taking limit N g ∞, we get probability of

**P(X U Y) = P(X) + P(Y)**

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

**1.** One card is drawn

from a well-shuffled deck of 52 cards. What is the probability of getting a

king or an ace?

**Solution:**

Let X be the event of ‘getting

a king’ and,

Y be the event of ‘getting

an ace’

We know that, in a well-shuffled deck of 52 cards there are 4

kings and 4 aces.

Therefore, probability

of getting a king from well-shuffled

deck of 52 cards = P(X) = 4/52 = 1/13

Similarly, probability of getting

an ace from well-shuffled deck of 52

cards = P(Y) = 4/52 = 1/13

According to the

definition of mutually exclusive we know that, drawing of a well-shuffled deck of 52 cards ‘getting

a king’ and ‘getting an ace’ are known as mutually exclusive events.

We have to find out

P(King or ace).

So according to the addition

theorem for mutually exclusive events, we get;

**P(X U Y) = P(X) + P(Y)**

Therefore, P(X U Y) | = 1/13 + 1/13
= (1 + 1)/13 = 2/13 |

Hence, probability of getting a king or an ace from a well-shuffled deck of 52 cards = 2/13

**2.** A bag contains 8 black pens and 2 red pens and if a pen is drawn at

random. What is the probability that it is black pen or red pen?

**Solution:**

Let X be the event of

‘getting a black pen’ and,

Y be the event of

‘getting a red pen’.

We know that, there are 8 black pens and 2 red pens.

Therefore, probability

of getting a black pen = P(X) = 8/10 = 4/5

Similarly, probability of getting a red pen = P(Y) = 2/10 = 1/5

According to the

definition of mutually exclusive we know that, the event of ‘getting a black pen’ and ‘getting a red pen’ from a bag are known as

mutually exclusive event.

We have to find out P(getting

a black pen or getting a red pen).

So according to the

addition theorem for mutually exclusive events, we get;

P(X U Y) = P(X) + P(Y)

Therefore, P(X U Y) | = 4/5 + 1/5
= 5/5 = 1 |

Hence, probability of getting ‘a black pen’ or ‘a red pen’ = 1