5.(a)  Out of  \small 100  students, two sections of \small 40 and \small 60 are formed. If you and your friend are among the \small 100 students, what is the probability that

(a) you both enter the same section? 

Answers (1)
H Harsh Kankaria

Total number of students = 100

Let A and B be the two sections consisting of 40 and 60 students respectively.

Number of ways of selecting 2 students out of 100 students.= ^{100}\textrm{C}_{n}

(a) 

If both are in section A:

Number of ways of selecting 40 students out of 100 = ^{100}\textrm{C}_{40}  (The remaining 60 will automatically be in section B!)

Remaining 38 students are to be chosen out of (100-2 =) 98 students

\therefore Required probability if they both are in section A = \frac{^{98}\textrm{C}_{38}}{^{100}\textrm{C}_{40}}

Similarly,

If both are in section B:

Number of ways of selecting 60 students out of 100 = = ^{100}\textrm{C}_{60} = ^{100}\textrm{C}_{40}  (The remaining 40 will automatically be in section A!)

Remaining 58 students are to be chosen out of (100-2 =) 98 students

\therefore  Required probability if they both are in section B = \frac{^{98}\textrm{C}_{58}}{^{100}\textrm{C}_{60}}

Required probability that both are in same section = Probability that both are in section A + Probability that both are in section B 

 = \frac{^{98}\textrm{C}_{38}}{^{100}\textrm{C}_{40}}+\frac{^{98}\textrm{C}_{58}}{^{100}\textrm{C}_{60}}               

 \\ = \frac{^{98}\textrm{C}_{38}+^{98}\textrm{C}_{58}}{^{100}\textrm{C}_{40}} \\ \\ =\frac{\frac{98!}{38!.60!} + \frac{98!}{58!.40!}}{\frac{100!}{40!.60!}}

=\frac{85}{165} = \frac{17}{33}

Hence, the required probability that both are in same section is \frac{17}{33}

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