PROBLEMS BASED ON MUTUALLY EXCLUSIVE EVENTS OR DISJOINT EVENTS

$ { P(A \cup B)=P(A)+P(B)}  $ (OR) $ { P(A + B)=P(A)+P(B)}$

EXAMPLE 1
One cards is drawn from a pack of 52 cards.What is the probability that it is either a king or a queen.

Solution   Given total number of  card is 52

     $ \therefore n(S)=52 $

Let A be a event that the card drawn is king

$ n(A)=4 $
$ P(A)=\frac{n(A)}{n(S)}=\frac{4}{52}=\frac{1}{13} $

Let B be a event that the card drawn is queen

$ n(B)=4 $
$ P(B)=\frac{n(B)}{n(S)}=\frac{4}{52}=\frac{1}{13} $

Here A  and B are disjoint events since both cannot occur together
.
$ P(A \cup  B) = P(A)+P(B) $        [  A   and    B  are   mutually exclusive   events]
$ = \frac{1}{13}+\frac{1}{13} $         [$ P(A)=\frac{1}{13} P(B)=\frac{1}{13}$]
$ P(A \cup  B) = \frac{2}{13}$

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EXAMPLE 2
From a group of 5 first year,4 second year and 4 third year students,3 students are selected at random.Find the probability that they are first year or third year students.

Solution: Given

First year      Second year      Third year       Total
5                     4                             4                  13

Three students are selected at random in total  number of students

$  n(S)= {}^{13}C_3=286 $       [ $  {}^nC_r=\frac{n!}{(n-r)!r!} $ ]

Let A be three students selected from the first year students

$ n(A) = {}^5C_3=10 $             [$ {}^nC_r=\frac{n!}{(n-r)!r!}$ ]   

$P(A) =  \frac{n(A)}{n(S)}=\frac{10}{286}$


Let B be three students selected from the third year students

$ n(B) = {}^4C_3=4 $          [$ {}^nC_r=\frac{n!}{(n-r)!r!}$ ]
$ P(B) = \frac{n(B)}{n(S)}=\frac{4}{286} $

Here A  and B are disjoint events since both cannot occur together.

$ P(A \cup  B) = P(A)+P(B) $         [  A  and   B   are mutually  exclusive    events]
                   $ = \frac{10}{286}+\frac{4}{286} $             [$ P(A)=\frac{10}{286} P(B)=\frac{4}{286}$ ]
$ P(A \cup  B)  =  \frac{14}{286} $
                   $ = \frac{7}{143} $




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