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} \ldots \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
$=1-\frac{17}{33}=\frac{16}{33}$