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Question
Out of 100 students, two sections of 40 and 60 are formed. If you and your friend are among the 100 students, what is the probability that
- you both enter the same sections?
- you both enter the different sections?
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Solution
Let there be two sections A and B having 40 and 60 students, respectively.
a. i. Suppose both the students belong to section A.
∴ 38 students are selected out of 98 students.
Ways to select 38 students from 98 = 98C38
Ways to select 40 students from 100, without any condition n(S) = 100C40
Probability that both students (he and his friend) get admission in the same section A
= `(""^98C_38)/(""^100C_40)`
= `(98!)/(38!60!) xx (40!60!)/(100!)`
= `(98! xx 40! xx 60!)/(38!60! xx 100.99(98!))`
= `(40.39)/(100 xx 99)`
= `26/165`
ii. If both students get admission in section B. Then number of ways to select 58 students from 98 = 98C58
number of ways to select 60 students from 100 = 100C60
So, if those students get admission in section B then the probability is
= `""^98C_58 ÷ ""^100C_60`
= `(98!)/(58!40!) ÷ (100!)/(60!40!)`
= `(98!)/(58!40!) xx (60 xx 59 xx (58!) xx (40!))/(100 xx 99 xx 98!)`
= `(60.59)/(100.99)`
= `59/(5 xx 33)`
= `59/165`
Both the students get admission in section A or B Then his probability
= `26/165 + 59/165`
= `85/165`
= `17/33`
b. Probability of both students getting admission in different sections
= `1 - 17/33`
= `(33 - 17)/33`
= `16/33`
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