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Question
Out of 100 students, two sections of 40 and 60 are formed. If you and your friend are among 100 students, what is the probability that: (i) you both enter the same section? (ii) you both enter the different sections?
Solution
(a ) when both enter the same section .
Here possibilities of two cases .
case 1 :- enter both are in section A
if both are in section A , 40 students out of 100 can be selected n ( S ) = ¹⁰⁰C₄₀
and ( 40 - 2) = 38 students out of ( 100 - 2) = 98 can be selected n ( E ) = ⁹⁸C₃₈
so, `P ( E ) = (n ( E ))/(n ( S ))`
= `(⁹⁸C₃₈) / (¹⁰⁰C₄₀)`
=` ((98!)/(38!) × 60! )/( (100!)/(40!) × 60! )`
= `((98!) × (40!) × 60!)/((38!) × (60!) × 100!)`
= `( 98!)/(100!) × ( 40)/(38!)`
=` 1/(100 × 99) × 40 × 39`
= `26/165`
case 2 :- if both are in section B, 60 students out 100 can be selected n( S )= ¹⁰⁰C₆₀
and (60 - 2) = 58 students out of ( 100 - 2)= 98 can be selected n( E ) = ⁹⁸C₅₈
so, P ( E ) =` (n ( E ))/(n (S ))`
= `(⁹⁸C₅₈) / (¹⁰⁰C₆₀)`
=` ((98!)/(58!) × 40! )/ ((100!)/(60!) × 40!)`
=` (98! × 60! × 40!)/(58! × 30! × 100!)`
= ` (98!)/(100!) × (60!)/(58!) × (40!)/(40!)`
= `{ 1/100 × 99 } × { 60 × 59 } × 1`
= `59/165`
Hence, Probability that students are either in section A or B .
= `26/ 165 + 59/165`
=` 85/165`
= `17/33`
(b ) We know,
P( E ) = 1 - P(E')
e.g
The Probability that both enter different sections = 1 - Probability that both enter same sections
= 1 - `17/33`
=` (33 - 17)/33`
= `16/33`
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