Question

Out of $100$ students, two sections of $40$ and $60$ are formed. Assume that you and your friend are among the $100$ students. Let $P\left(A\right)$ be the probability that you both enter in same section and $P\left(B\right)$ be the probability that you both enter in different sections. Then which of the following is/are true?

• A. $P\left(B\right)=\frac{17}{33}$
• B. $P\left(B\right)=\frac{16}{33}$
• C. $P\left(A\right)=\frac{17}{33}$
• D. $P\left(A\right)=\frac{16}{33}$

Answer 1

Answer: (B), (C)

$P\left(A\right)=\frac{17}{33}$

Sample space will be ${}^{100}{C}_{40}{\times }^{60}{C}_{60}\text{or}{}^{100}{C}_{60}{\times }^{40}{C}_{40}$ (we can either select $40$ students first or $60$ students first)

$A$ : They both will enter the same section if both are among $40$ or both are among $60$
$\therefore$ Number of favourable ways $={}^2{C}_2\times {}^{98}{C}_{38}+{}^2{C}_2\times {}^{98}{C}_{58}$
$\therefore P\left(A\right)=\frac{{}^2{C}_2\times {}^{98}{C}_{38}+{}^2{C}_2\times {}^{98}{C}_{58}}{{}^{100}{C}_{40}{\times }^{60}{C}_{60}}$
$\Rightarrow P\left(A\right)=\frac{\frac{(98!)}{38!60!}+\frac{(98!)}{58!40!}}{\frac{(100!)}{40!60!}}=\frac{17}{33}$
$\Rightarrow P\left(A\right)=\frac{(39\times 40)+(59\times 60)}{100\times 99}=\frac{17}{33}$

$\left(B\right)$ : Both will enter two different sections
$P\left(B\right)=1-p$ (both enter same section)
$\therefore P\left(B\right)=1-\frac{17}{33}=\frac{16}{33}$