2020-06-11T17:59:05-04:00
The following is a breakdown of faculty members at IIM-A.
Rank No of males No of females
Professor 20 12
Associate Prof 18 20
Asst Prof 25 30.
What is the probability that a faculty member selected at random is
a) Female.
b) A female professor
c) A female or professor.
d)A professor or asst. Prof
1
2020-06-14T18:31:31-0400
M a l e F e m a l e T o t a l P r o f e s s o r 20 12 32 A s s o c i a t e P r o f 18 20 38 A s s i s t a n t P r o f 25 30 55 T o t a l 63 62 125 \def\arraystretch{1.5}
\begin{array}{c:c:c:c}
& Male & Female & Total \\ \hline
Professor & 20 & 12 & 32 \\
\hdashline
Associate\ Prof & 18 & 20 & 38 \\
\hdashline
Assistant \ Prof & 25 & 30 & 55 \\
\hdashline
Total & 63 & 62 &125
\end{array} P ro f essor A ssoc ia t e P ro f A ss i s t an t P ro f T o t a l M a l e 20 18 25 63 F e ma l e 12 20 30 62 T o t a l 32 38 55 125 a)
P ( F ) = 62 125 = 0.496 P(F)={62\over 125}=0.496 P ( F ) = 125 62 = 0.496 b)
P ( F ∩ P r o f ) = 12 125 = 0.096 P(F\cap Prof)={12\over 125}=0.096 P ( F ∩ P ro f ) = 125 12 = 0.096 c)
P ( F ∪ P r o f ) = P ( F ) + P ( P r o f ) − P ( F ∩ P r o f ) = P(F\cup Prof)=P(F)+P(Prof)-P(F\cap Prof)= P ( F ∪ P ro f ) = P ( F ) + P ( P ro f ) − P ( F ∩ P ro f ) =
= 62 125 + 32 125 − 12 125 = 82 125 = 0.656 ={62\over 125}+{32\over 125}-{12\over 125}={82\over 125}=0.656 = 125 62 + 125 32 − 125 12 = 125 82 = 0.656 d)
P ( P r o f ∪ A s s i s t P r o f ) = P ( F ) + P ( A s s i s t P r o f ) = P(Prof\cup Assist\ Prof)=P(F)+P(Assist \ Prof)= P ( P ro f ∪ A ss i s t P ro f ) = P ( F ) + P ( A ss i s t P ro f ) =
= 32 125 + 55 125 = 87 125 = 0.696 ={32\over 125}+{55\over 125}={87\over 125}=0.696 = 125 32 + 125 55 = 125 87 = 0.696
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