Question

If the papers of 4 students can be checked by any one of the 7 teachers, then the probability that all the 4 papers are checked by exactly 2 teachers is
(a)$\dfrac{2}{7}$
(b)$\dfrac{6}{49}$
(c)$\dfrac{32}{343}$
(d)None of these

Answer 1

Hint: Find the total number of ways of checking 4 papers by 7 teachers. Each teacher may check all four. Now find the number of ways of choosing 2 teacher’s out of 7, cancel the chances of 1 teacher checking the paper. Hence favorable conditions will be their product and thus find the probability.

Complete step-by-step answer:
We have been given that there are 4 students and 7 teachers checking their papers. So, we can write that the
Total number of papers = 4
Total number of teachers = 7
Thus we can get the total number of ways in which the papers of 4 students can be checked by 7 teachers as $7\times 7\times 7\times 7$ , i.e. each paper can be checked by 7 teachers. Hence the 4 papers can be checked by any of the 7 teachers and it’s not mentioned that they are not recurring.
$\therefore$ Total number of ways of checking papers = $7\times 7\times 7\times 7={{7}^{4}}$.
We can write the number of ways of choosing 2 teachers out of 7 = ${}^{7}{{C}_{2}}$.
We use combination as ordering is not important here. So the formula is given as

& {}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!} \\\ & \therefore {}^{7}{{C}_{2}}=\dfrac{7!}{2!\left( 7-2 \right)!}=\dfrac{7\times 6\times 5!}{5!\times 2}=21 \\\ \end{aligned}$$ $$\therefore $$ We get the number of ways of choosing 2 teachers out of 7 = $${}^{7}{{C}_{2}}=21$$. Now, let us consider the number of ways in which there 2 teachers can check 4 papers. Then, we will get it as $$2\times 2\times 2\times 2={{2}^{4}}$$ , i.e. each paper can be checked by 2 teachers and it can be recurring. But this might include two ways in which all the paper will be checked by a single teacher. $$\therefore $$ We get the number of ways in which 4 papers can be checked by exactly 2 teachers = $${{2}^{4}}$$ - 2 = 16 – 2 = 14. Hence, we can write the number of favorable ways of checking the paper = $${}^{7}{{C}_{2}}\times 14=21\times 14$$. $$\therefore $$ Required probability = Number of favorable ways / Total number of ways. $$\therefore $$ Required probability = $$\dfrac{21\times 14}{{{7}^{4}}}=\dfrac{21\times 14}{7\times 7\times 7\times 7}=\dfrac{3\times 2}{7\times 7}=\dfrac{6}{49}$$. Hence we got the probability as $$\dfrac{6}{49}$$. $$\therefore $$ Option (b) is the correct answer. Note: You can first mistakenly take the total number of ways in which papers of 4 students checked by 7 teachers as $${}^{7}{{C}_{4}}$$. This will provide you with the wrong answer. The 4 papers can be checked by any one of the 7 teachers. There are chances that even one teacher can check all the 4 papers.