Question

Out of $100$ students, two sections of $40$ and $60$ are formed. If you and your friend are among the $100$ students.
Statement I The probability that you both enter the same section, is $\frac{17}{33}$.
Statement II The probability that you both enter the different sections, is $\frac{16}{33}$

• A. Statement I is true; Statement II is true; Statement II is a correct explanation for Statement I
• B. Statement I is true; Statement II is true; Statement II is not a correct explanation for Statement I
• C. Statement I is true; Statement II is false
• D. Statement I is false; Statement II is true

Answer 1

Answer: (B)

Let the two sections have $A$ and $B$ are of $40$ and $60$ students respectively.
A. When both enter the same section There are two cases,
Case I Either both are in section $A$ or
Case II Either both are in section $B$.
I. If both are in section $A$, then $40$ students out of $100$ students can be selected $={}^{100}{C}_{40}$ ways
and $38$ students can be selected out of $98$ students
$={}^{98}{C}_{38}$ ways
$\therefore$ Probability $P=\frac{{}^{98}{C}_{38}}{{}^{100}{C}_{40}}$
$=\frac{98!}{38!60!}\times \frac{40!60!}{100!}\left(\because {}^n{C}_r=\frac{n!}{r!(n-r)!}\right)$
$=\frac{98!\times 40\times 39\times 38!}{38!\times 100\times 99\times 98!}$
$=\frac{40\times 39}{100\times 99}=\frac{2}{5}\times \frac{13}{33}=\frac{26}{165}\dots \text{ (i) }$
II. If both are in section $B$. Then, $60$ students out of $100$ students can be selected $={}^{100}{C}_{60}$ ways
$58$ students out of $98$ students can be selected
$={}^{98}{C}_{58}$
$\therefore$ Probability $=\frac{{}^{98}{C}_{58}}{{}^{100}{C}_{60}}=\frac{98!}{58!40!}\times \frac{60!40!}{100!}$
$=\frac{98!\times 60\times 59\times 58!\times 40!}{58!\times 40!\times 100\times 99\times 98!}$
$=\frac{60\times 59}{100\times 99}=\frac{3}{5}\times \frac{59}{99}=\frac{59}{5\times 33}=\frac{59}{165}.....$(ii)
Hence, required probability that students are either in section $A$ or $B$
$=\frac{26}{165}+\frac{59}{165}$ (from Eqs. (i) and (ii))
$=\frac{85}{165}=\frac{17}{33}$
(b) The probability that both enter the different section $=1$ - Probability that they both enter the same section
$=1-\frac{17}{33}=\frac{16}{33}$