Question

10 different books and 2 different pens are given to 3 boys so that each gets an equal number of things. The probability that the same boy does not receive both the pens is
(A) $\dfrac{5}{{11}}$
(B) $\dfrac{7}{{11}}$
(C) $\dfrac{2}{3}$
(D) $\dfrac{8}{{11}}$

Answer 1

Hint :We will use this property of probability in above question
Probability of occurring of event $ = 1 - $ probability of not occurring of event.
Then we will solve the question from the combination formula.

Formula used -
Combination formula $^n{C_r} = \dfrac{{n!}}{{r!(n - r)!}}$
Where
n $ = $ total number of items
r $ = $ number of items to be selected from sample
Symbol represents factorial
Here $n! = n \times (n - 1) \times (n - 2) \times (n - 3) \times ..... \times 1$
and $r! = r \times (r - 1) \times (r - 2) \times (r - 3) \times ..... \times 1$


Complete step by step solution :
Probability of the same boy does not receive both pens $ = 1 - $ probability of any boy receiving both pens.
Since, all the students receives same number of articles, so each on gets $ = \dfrac{{(10 + 2)}}{3} = 4$ items
Number of ways in which 10 books and 2 pens is distributed to 3 students
${ = ^{12}}{C_4}{ \times ^8}{C_4}{ \times ^4}{C_4}$
Now, to find number of ways such that any boy receive both pens
Selection of a boy from 3 boys ${ = ^3}{C_1}$
Giving both the pens to him ${ = ^3}{C_1}{ \times ^2}{C_2}$
$ = {(^3}{C_1} \times 1)$ ways
He got two items
Ways of giving rest of items to him ${ = ^{10}}{C_2}$ ways
and now we have 8 items. Other two boys get things in $ = {(^8}{C_4}{ \times ^4}{C_4})$ ways
So number of ways such that any boy receive both pens \[ = {(^3}{C_1} \times 1) \times {(^{10}}{C_2}) \times {(^4}{C_4}{ \times ^4}{C_4})\]
Probability of any boy get both pens, ${P_1} = \dfrac{{^3{C_1}{ \times ^{10}}{C_2}{ \times ^8}{C_4}{ \times ^4}{C_4}}}{{total\,number\,of\,ways}}$
$ = \dfrac{{^3{C_1}{ \times ^{10}}{C_2}{ \times ^8}{C_4}{ \times ^4}{C_4}}}{{^{12}{C_4}{ \times ^8}{C_4}{ \times ^4}{C_4}}}$
$ = \dfrac{{3 \times \dfrac{{10 \times 9}}{{1 \times 1}}}}{{\dfrac{{12 \times 11 \times 10 \times 9}}{{4 \times 3 \times 2 \times 1}}}}$
$ = \dfrac{{3 \times 10 \times 9}}{{2 \times 1}} \times \dfrac{{4 \times 3 \times 2 \times 1}}{{12 \times 11 \times 10 \times 9}}$
${P_1} = \dfrac{3}{{11}}$
Required probability $ = 1 - {P_1}$
$ = 1 - \dfrac{3}{{11}}$
$P = \dfrac{8}{{11}}$
So, the probability that the same boy does not receive both the pens is $\dfrac{8}{{11}}$
Therefore, option D i.e., $\dfrac{8}{{11}}$ is the correct option.


Note : Here in solution we have used a short trick to find out combination values instead of using formula.
Here is that trick.
$^5{C_3} = ?$
Method 1
$^n{C_r} = \dfrac{{n!}}{{r!(n - r)!}}$
$n = 5$, $r = 3$
$^5{C_3} = \dfrac{{5!}}{{3!(5 - 3)!}} = \dfrac{{5 \times 4 \times 3 \times 2 \times 1}}{{3 \times 2 \times (2)!}}$
$ = 10$
Method 2 (Short method)
$^5{C_3} = \dfrac{{5 \times 4 \times 3}}{{3!}}$
$ = \dfrac{{5 \times 4 \times 3}}{{3 \times 2}}$
$ = 10$