Mutually Exclusive Events

• In throwing a die all the $6$ faces numbered $1$ to $6$ are mutually exclusive since if any one of these faces comes, the possibility of others, in the same trial, is ruled out.

• Similarly in tossing a coin the events head and tail are mutually exclusive.

• In throwing a die, let $A$ be the event of getting an odd number and $B$ be the event of getting an even number. Then,

$A = \1, 3, 5\$ and $B = \2, 4, 6\$.

The events $A$ and $B$ are mutually exclusive, since when we throw a die either an even number or an odd number shows up but not both.

Now,

Exhaustive number of cases $n(\Omega) = 6,$

Number of cases favourable to $A$ is $n(A) = 3,$

Number of cases favourable to $B$ is $n(B) = 3.$

$\therefore \quad$Probability of getting an odd number is

$P(A) = \fracn(A)n(\Omega) = \frac36 = \frac12.$

Probaility of getting an even number is

$P(B) = \fracn(B)n(\Omega) = \frac36 = \frac12.$

It is clear that

$A \cup B = \1, 2, 3, 4, 5, 6\ = \Omega,$ and

$P(A) + P(B) = \frac12 + \frac12 = 1 = P(\Omega) = P(A \cup B).$

• In the random experiment of throwing two dice, let $X$ be the event of getting a sum of $8$ and $Y$ be the event of getting a sum of $6$. Then,

$X = \(2, 6), (3, 5), (4, 4), (5, 3), (6, 2)\,$

$Y = \(1, 5), (2, 4), (3, 3), (4, 2), (5, 1)\.$

Here, the events $X$ and $Y$ are mutually exclusive.

Now,

Exhaustive number of cases $n(\Omega) = 36,$

Number of favourable cases to $X$ is $n(X) = 5,$

Number of favourable cases to $Y$ is $n(Y) = 5.$

$\therefore \quad$Probability that the total of numbers on the two dice to be $8$ is

$P(X) = \fracn(X)n(\Omega) = \frac536.$

Probability that the total of numbers on the two dice to be $6$ is

$P(Y) = \fracn(Y)n(\Omega) = \frac536.$

Now, let us consider an event $Z$ of getting either $6$ or $9$ as the total of numbers on the two dice. Then

$Z = \(1, 5), (2, 4), (2, 6), (3, 3), (3, 5), (4, 2), (4, 4), (5, 1), (5, 3), (6, 2)\.$

Number of cases favourable to $Z$ is $n(Z) = 10.$

$\therefore \quad$Probability that the total of numbers on the two dices is either $6$ or $8$ is

$P(Z) = \fracn(C)n(\Omega) = \frac1036 = \frac518.$

It is clear that

$X \cup Y = Z$, and

$P(X) + P(Y) = \frac536 + \frac536 = \frac1036 = \frac518 = P(Z) = P(X \cup Y).$

From the above discussion it is clear that

probability Mutually Exclusive Events , what are Mutually Exclusive Events in probability concept