Explain Solution R.D. Sharma Class 12 Chapter 30 Probability  Exercise 30.5 Question 34 Sub Question 1 Maths Textbook Solution.

Answers (1)

Answer:\frac{17}{33}

Hint: You must know the rules of finding probability functions.

Given: Out of 100 students, two section of 40 and 60 are formed

Solution: When both enter the same section.

Here are the possibilities of two cases:

Case-(1)\rightarrowenter both are in section A

If both are in section A, 40 students out of 100 can be selected n(s)={ }^{100} C_{40}

and (40-2)=38 students out of (100-2)=98

Can be selected n(E)={ }^{98} C_{38}

So,

\begin{aligned} &P(E)=\frac{n(E)}{n(s)}=\frac{{ }^{98} C_{38}}{{ }^{100} C_{40}} \\ &=\frac{98 !}{38 !} \times \frac{40 !}{100 !} \times \frac{60 !}{60 !} \\ &=\frac{1}{100 \times 99} \times 40 \times 39 \\ &=\frac{26}{165} \end{aligned}

Case-(2)\rightarrowif both are in section B, 60 students out of 100 can be selected n(s)={ }^{100} C_{60}

and (60-2)=58 students out of (100-2)=98

Can be selected { }^{98} C_{58}

So,

\begin{aligned} &P(E)=\frac{n(E)}{n(s)}=\frac{{ }^{98} C_{58}}{{ }^{100} C_{60}} \\ &=\frac{98 !}{58 !} \times \frac{60 !}{100 !} \times \frac{40 !}{40 !} \\ &=\left\{\frac{1}{100} \times 99\right\} \times\{60 \times 59\} \times 1 \\ &=\frac{59}{165} \end{aligned}

Hence, probability that student are either in sector

    \begin{aligned} &=\frac{26}{165}+\frac{59}{165} \\ &=\frac{26+59}{165}=\frac{85}{165}=\frac{17}{33} \end{aligned}

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