Question

There are 10 students of which three are graduates. If a committee of five is to be formed,
what is the probability that there are (1) only 2 graduates (i) atleast 2 graduates? ​

Answer 1

$\displaystyle\huge\red{\underline{\underline{Solution}}}$

GIVEN

• There are 10 students
• Among them three are graduates
• A committee of five is to be formed

TO DETERMINE

The probability that there are

1. only 2 graduates
2. Atleast 2 graduates

CALCULATION

Total number of students = 10

Among them three are graduates

A committee of five is to be formed

So 5 students can be selected from 10 students in

$= \sf{\large{ {}^{10} C_5} \: } = 252 \: \: \: ways$

So the total number of possible outcomes = 252

ANSWER TO QUESTION : 1

Now 2 graduates can be selected from 3 graduates in

$= \sf{\large{ {}^{3} C_2} \: } = 3 \: \: \: ways$

In order to make a committee of 5 students rest ( 5- 2) = 3 students can be selected from rest (10-3)= 7 students in

$= \sf{\large{ {}^{7} C_3} \: } = 35 \: \: \: ways$

Let A be the event that there are only 2 graduates

So the total number of possible outcomes for the event A is = 35 × 3 = 105

So the required probability

$\displaystyle \: = \sf{P(A) = \frac{105}{252} = \frac{5}{12} }$

ANSWER TO QUESTION : 2

Now atleast 2 graduates are to be selected.

So the number of graduates is either 2 or 3

Now 2 graduates can be selected from 3 graduates in

$= \sf{\large{ {}^{3} C_2} \: } = 3 \: \: \: ways$

In order to make a committee of 5 students rest ( 5- 2) = 3 students can be selected from rest (10-3)= 7 students in

$= \sf{\large{ {}^{7} C_3} \: } = 35 \: \: \: ways$

So the total number of ways in this case = 35 × 3 = 10

Again

3 graduates can be selected from 3 graduates in

$= \sf{\large{ {}^{3} C_3} \: } = 1 \: \: \: ways$

In order to make a committee of 5 students rest ( 5- 3) = 2 students can be selected from rest (10-3)= 7 students in

$= \sf{\large{ {}^{7} C_2} \: } = 21 \: \: \: ways$

So the total number of ways in this case = 21 × 1 = 21

Let B be the event that Atleast 2 graduates are to be selected

Then the total number of possible outcomes for the event B is = 105 + 21 = 126

So the required probability is

$\displaystyle \: = \sf{P(B) = \frac{126}{252} = \frac{1}{2} }$

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