Question

Out of 100 students, two sections of 40 and 60 are formed. If you and your friend are among the 100 students and let
E : event that you both enter the same section and
F : event that you both enter the different section, then

• A. $P(E)+P(F)=1$
• B. E and F are equally likely and exhaustive.
• C. $P(E)>P(F)$
• D. $P(E)<P(F)$

Answer 1

Answer: (A), (C)

$n(S)=\frac{100!}{40!\cdot 60!}$
$n(F)=\frac{98!}{39!\cdot 59!}\cdot 2!\text{ (think!) }$
$P(F)=\frac{98!\cdot 2!}{39!\cdot 59!}\frac{40!\cdot 60!}{100!}=\frac{2\cdot 40\cdot 60}{100\cdot 99}=\frac{48}{99}=\frac{16}{33}$
$P(E)=\frac{16}{33};P(E)=\frac{17}{33}\Rightarrow (A)/(C)$ are correct.