Exclusive Events Probability

In probability theory, events E1, E2, ..., En are said to be mutually exclusive if the occurrence of any one of them automatically implies the non-occurrence of the remaining n − 1 events. Therefore, two mutually exclusive events cannot both occur. Mutually exclusive events have the property: P (A ∩ B) = 0.When A and B are mutually exclusive, P (A or B) = P (A) + P (B).


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Conditional probability for mutually exclusive events

If, A nn B = O/  two events are mutually exclusive

Then we know that p ( A / B) = p( A   nn   B  ) / P ( B)

Then if P ( B ) = 0

Then P ( A / B ) = 0.

Example problem for Mutually Exclusive Events:

Pro 1: What is the probability of die showing 2 or 5?

Sol :        P (2) = 1/6

P (5) = 1/6

P (2 nn 5) = P (2) + P (5)

= (1/6) + (1/6)

= 2/6

P (2 nn 5)   = 1/3

The Probability of a die showing 2 or 5 is 1/3

Pro 2:   The probability of the teams P, Q and R is winning a badminton competition are (2/4).(2/8) and (2/10) respectively.

Calculate the probability that

a) Either P or Q will win

b) Either P or Q or R will win

c) None of these teams will win

d) Neither P nor Q will win

Sol :       a) P(P or Q will win)

= (2/4)+(2/8)

= (6/8)

P(P or Q will win)     = (3/4)

b) P (P or Q or R will win)

= (2/4) + (2/8) + (2/10)

= (160/320)

P (P or Q or R will win)   = (3 / 2)

c) P (none will win)

= 1 – P (P or Q or R will win)

= 1- (3 / 2)

P (none will win)  = (- 1 / 2)

d) P (neither P nor Q will win)

= 1 – P(either P or Q will win)

= 1 - (3 / 2)

P (neither P nor Q will win)   = (-1 / 2)


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Pro 3:   What is the probability to draw a king, queen, or jack from a deck of playing cards?

Sol :        The individual probabilities are

King = (4/52)

Queen = (4/52)

Therefore, the probability of success is

P = (4/52)+(4/52)

= (8/52)

P = (2/13)