$\begin{array}{1 1}(A)\;\large\frac{16}{33}\\(B)\;\large\frac{17}{33}\\(C)\;\large\frac{36}{165}\\(D)\;\large\frac{56}{165}\end{array} $

- $nC_r=\large\frac{n!}{r!(n-r)!}$
- Required probability=$\large\frac{n(E)}{n(S)}$

Step 1:

Both enter the different section

$\therefore$ Probability that enter different section =1-probability that both enter same section.------(1)

If both are in section A then 40 students out of 100 students will be selected.

Since both of them are together,rest 38 students out of 98 students are chosen

$\therefore$ Required probability =$\large\frac{n(E)}{n(S)}$

$\Rightarrow \large\frac{98C_{38}}{100C_{40}}$

$\Rightarrow \large\frac{\Large\frac{98!}{38!(98-38)!}}{\Large\frac{100!}{40!(100-40)!}}$

$\Rightarrow \large\frac{40\times 39}{100\times 99}$

$\Rightarrow \large\frac{26}{165}$

$\therefore$ The probability when both are in section A =$\large\frac{26}{165}$

Step 2:

Both are in section B

If both are in section B then 60 students out of 100 students will be selected.

Since both of them are together,rest 58 students out of 98 students are chosen

$\therefore$ Required probability =$\large\frac{n(E)}{n(S)}$

$\Rightarrow \large\frac{98C_{58}}{100C_{60}}$

$\Rightarrow \large\frac{\Large\frac{98!}{58!(98-58)!}}{\Large\frac{100!}{60!(100-60)!}}$

$\Rightarrow \large\frac{98!}{58!\times 40!}\times \frac{60!\times 40!}{100!}$

$\Rightarrow \large\frac{59}{165}$

$\therefore$ The probability when both are in section B =$\large\frac{59}{165}$

Step 3:

The probability that students are in either section A or section B=$\large\frac{26}{165}+\frac{59}{165}$

$\Rightarrow \large\frac{85}{165}=\frac{17}{33}$

$\therefore$ The probability that both enter the same section is $\large\frac{17}{33}$

Substituting this in (1) we get

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

$\therefore$ Probability that both enter different section is $\large\frac{16}{33}$

Hence (A) is the correct answer.

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