ISRO2018-14

Question

A class of $30$ students occupy a classroom containing $5$ rows of seats, with $8$ seats in each row. If the students seat themselves at random, the probability that sixth seat in the fifth row will be empty is:

1. $1/5$
2. $1/3$
3. $1/4$
4. $2/5$

Answer 1

The class contains $5$ rows of seats & each row having $8$ seats

∴ The classroom have a total of $(5 \times 8) = 40$ seats.

& the class has a total of $30$ students.

Now, the $6^{th}$ seat of row no. $5$ has to be empty.

& We need to find the probability of the same.

We know, $\text{Probability} = \dfrac{\text{No. of favorable ways}}{\text{Total no. of ways}}$

Now, the favorable ways will be when

$30$ students have the option to seat in $39$ seats. (Because $6^{th}$ seat of row no. $5$ needs to be empty)

∴ Out of $39$ seats $30$ students can seat in $^{39}C_{30}$ ways. [∵ As $30$ students have to choose $30$ seats out of $39$ seats.]

And Total no. of ways will be when

$30$ students have the option to seat in $40$ seats.

∴ No. of ways will be = $^{40}C_{30}$ways

$\color{violet}{\text{Now, the probability of the}}$ $\color{violet}{6^{th}}$ $\color{violet}{\text{ no. seat of row no.}}$ $\color{violet}{5}$ $\color{violet}{\text{ will left vacant is}}$ = $\dfrac{^{39}C_{30}}{^{40}C_{30}}$

$^{39}C_{30} = \dfrac{39 \times 38 \times .... \times 1}{(30 \times 29 \times .... \times 1) \times (9 \times 8 \times .... \times 1)}$

$\qquad \qquad = \dfrac{39 \times 38 \times .... \times 31}{9 \times 8 \times .... \times 1}$

$^{40}C_{30} = \dfrac{40 \times 39 \times .... \times 1}{(30 \times 29 \times .... \times 1) \times (10 \times 9 \times 8 \times .... \times 1)}$

$\qquad \qquad = \dfrac{40 \times 39 \times .... \times 31}{10 \times 9 \times 8 \times .... \times 1}$

∴$\dfrac{^{39}C_{30}}{^{40}C_{30}} =\dfrac{39 \times 38 \times .... \times 31}{9 \times 8 \times .... \times 1} \times \dfrac{10 \times 9 \times 8 \times .... \times 1}{40 \times 39 \times .... \times 31} $

$\qquad \qquad = \dfrac{10}{40}$

$\qquad \qquad = \color{maroon}{\dfrac{1}{4}}$

Comments

Hats off to your dedication.

Answer 2

There are a total of $8\times 5=40$ seats.

Since students take seats randomnly, the probability of a given student taking a particular seat is $\dfrac{1}{40}.$

As there are $30$ students so each one of them could take the particular seat and probability of this will be

$$30 . \dfrac{1}{40} = \dfrac{3}{4}.$$

So probability of any particular seat being empty $=1-\dfrac{3}{4} =\dfrac{1}{4}.$

Option C is the answer

Answer 3

There are total 5 row and 8 seat in each row.

Total seat=>5*8=40. 6th seat in 5 row need to be blank, so we have 39 for 30 seat.

no. of ways to choice 30 seat out 39 =>  $\binom{39}{30}$

Total ways to choice 30 seat of out 40=> $\binom{40}{30}$

Probability the 6th seat of 5th row left blank => $\frac{\binom{39}{30}}{\binom{40}{30}}$

ans. 1/4