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){\text{ }}\dfrac{5}{{11}} \\
  (b){\text{ }}\dfrac{7}{{11}} \\
  (c){\text{ }}\dfrac{2}{3} \\
  (d){\text{ }}\dfrac{8}{{11}} \\
$

Answer 1

Hint – In this question first find out the total number of items that will be given to a single individual as it will be the ratio of total number of items to total number of boys. Then check for the ways in which these total items to distribute these out of total items to a single boy using the concept of combination. This will help to approach the solution of this problem.

Complete step-by-step answer:
Total number of different books = 10
And total number of different pens = 2
So the total number of items = 10 + 2 =12
Now it is given that there are 3 boys and each get equal items.
So each boy gets $\dfrac{{12}}{3} = 4$ items.
Now first find the total number of ways so that each boy receives 4 items.
So the total number of ways to distribute 4 items out of 12 for first boy = ${}^{12}{C_4}$
Now there are 8 items remaining so the total number of ways to distribute 4 items out of 8 for the second boy = ${}^8{C_4}$
Now there are 4 items remaining so the total number of ways to distribute 4 items out of 4 for the third boy = ${}^4{C_4}$
So the total number of ways to distribute 12 items so that each boy received 4 items =${}^{12}{C_4}{}^8{C_4}{}^4{C_4}$.
So the total number of outcomes =${}^{12}{C_4}{}^8{C_4}{}^4{C_4}$.
Now first assume that the same boy receives both the pen so the number of ways to choose that boy = ${}^3{C_1}$
Now the same boy deserves two more items from the remaining (12 – 2) =10 items
So the number of ways to distribute 2 item out of 10 for the same boy = ${}^{10}{C_2}$
So the number of ways for the same boy to get 2 different pens and 2 different books = ${}^3{C_1} \times {}^{10}{C_2}$
Now the remaining items are 8 different books.
So the number of ways to distribute 4 items equally to the rest of the boys = ${}^8{C_4}{}^4{C_4}$.
So the total number of ways that same student get 2 pens = ${}^3{C_1} \times {}^{10}{C_2} \times {}^8{C_4}{}^4{C_4}$
So the probability (P) that the same student gets 2 pens is the ratio of total number of outcomes to the favorable number of outcomes.
$ \Rightarrow p = \dfrac{{{}^3{C_1} \times {}^{10}{C_2} \times {}^8{C_4}{}^4{C_4}}}{{{}^{12}{C_4}{}^8{C_4}{}^4{C_4}}} = \dfrac{{{}^3{C_1} \times {}^{10}{C_2}}}{{{}^{12}{C_4}}}$
Now we have to find the probability so that the same boy does not receive both the pens.
So this probability (${p_1}$) is equal to difference of 1 and above probability as total probability is always 1.
$ \Rightarrow {p_1} = 1 - p = 1 - \dfrac{{{}^3{C_1} \times {}^{10}{C_2}}}{{{}^{12}{C_4}}}$
Now as we know that ${}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$ so use this property in above equation we have,
$ \Rightarrow {p_1} = 1 - \dfrac{{{}^3{C_1} \times {}^{10}{C_2}}}{{{}^{12}{C_4}}} = 1 - \dfrac{{3 \times \dfrac{{10!}}{{2!8!}}}}{{\dfrac{{12!}}{{4!8!}}}}$
Now simplify this we have,
$ \Rightarrow {p_1} = 1 - \dfrac{{3 \times \dfrac{{10!}}{{2!8!}}}}{{\dfrac{{12!}}{{4!8!}}}} = 1 - \dfrac{{3 \times 10!4!}}{{2!12!}} = 1 - \dfrac{{3 \times 10! \times 4 \times 3 \times 2!}}{{2! \times 12 \times 11 \times 10!}}$
$ \Rightarrow {p_1} = 1 - \dfrac{{3 \times 10! \times 4 \times 3 \times 2!}}{{2! \times 12 \times 11 \times 10!}} = 1 - \dfrac{3}{{11}} = \dfrac{{11 - 3}}{{11}} = \dfrac{8}{{11}}$
So this is the required probability.
Hence option (D) is the correct answer.

Note – The trick point here was that distribution was taken into consideration in such a way that the same boy receives both the pens. As we know that probability always lies between 0 to 1 that is $0 \leqslant P \leqslant 1$. Thus a simple subtraction of the value of this event’s probability when both the pens are given to a single boy with 1, helps obtain the probability of the event that when a single boy doesn't receive both the pens.