Question

A teacher-student committee consisting of 4 people is to be formed from 5 teachers and 20 students. Find the probability that the committee will consist of at least one student.

Answer 1

To find :-

Probability of forming a committee having at least 1 student.

Given :-

• Total required members in committee = 4

• Total teachers we have= 5

• Total students we have = 20

Solution :-

Firstly we'll find the Favourable number of ways in which the committee can be formed .

We need at least 1 student in our committee.

→ 3 teachers and 1 student or 2 teachers and 2 students or 1 teacher and 3 students or 4 students.

So we'll apply combination here :-

$\sf{\implies ^{5} C_3 \times ^{20} C_1 + ^{5} C_{2} \times ^{20} C_2 + ^{5} C_1 \times ^{20} C_3 + ^{20} C_4 }$

$\sf{ \implies \: (\frac{5 \times 4 \times 3}{3 \times 2 \times 1} \times \frac{20}{1} ) + ( \frac{5 \times 4}{2 \times 1} \times \frac{20 \times 19}{2 \times 1} ) + ( \frac{5}{1} \times \frac{20 \times 19 \times 18}{3 \times 2 \times 1} ) + ( \frac{20 \times 19 \times 18 \times 17}{4 \times 3 \times 2 \times 1} )}$

Now solving this fraction

$\sf{ \implies \: ( 5 \times 2 \times 20) +( 5 \times 2 \times 10 \times 19) + (5 \times 10 \times 19 \times 6) + 5 \times 19 \times 3 \times 17)}$

$\sf{\implies 10 + 380 + 1140 + 2295 }$

${\underline{\underline{\sf{\implies Favourable\: ways \: =\: 3825}}}}$

Now we'll find out the total ways of forming committee :-

Students + teacher = 20+5 → 25

$\sf{ \implies \: ^{25} C_5 = \frac{25 \times 24 \times 23 \times 22 \times 21}{5 \times 4 \times 3 \times 2 \times 1} }$

${\underline{\underline{\sf{\implies Total\:ways \: = \: 53130}}}}$

Now we have to find out the probability.

$\sf{\implies Probability \: P(A) \: = \dfrac{Favourable\: ways }{Total \: ways } }$

$\sf{\implies \: P(A) \: = \dfrac{3825}{53130 } }$

$\sf{\implies \: P(A) \: = \dfrac{765}{10626 } }$

$\sf{\implies \: P(A) \: = \dfrac{255}{3542} }$

${\underline{\underline{\sf{\implies \: Answer \: = \dfrac{255}{3542 } }}}}$