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 section?
Answers
Which team would they be on if they where on the same team, the numbers aren’t equal......sorry cant answer this my self but i can google it, just a min.......
5 HOURS LATER(not literally)
By: foxsrule
Have a foxy day!
Answer:
Step-by-step explanation:
(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