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.
$$\frac{2}{7}$$
B.
$$\frac{{12}}{{49}}$$
C.
$$\frac{{32}}{{343}}$$
D.
none of these
Answer :
none of these
Solution :
The total number of ways in which papers of $$4$$ students can be checked by seven teachers is $${7^4}.$$
The number of ways of choosing two teachers out of $$7$$ is $${}^7{C_2}.$$
The number of ways in which they can check four papers is $${2^4}.$$
But this includes two ways in which all the papers will be checked by a single teacher. Therefore, the number of ways in which $$4$$ papers can be checked by exactly two teachers is $${2^4} - 2 = 14.$$
Therefore, the number of favorable ways is $$\left( {{}^7{C_2}} \right)\left( {14} \right) = \left( {21} \right)\left( {14} \right).$$
Thus, the required probability is $$\frac{{\left( {21} \right)\left( {14} \right)}}{{{7^4}}} = \frac{6}{{49}}.$$
Releted MCQ Question on Statistics and Probability >> Probability
Releted Question 1
Two fair dice are tossed. Let $$x$$ be the event that the first die shows an even number and $$y$$ be the event that the second die shows an odd number. The two events $$x$$ and $$y$$ are:
Two events $$A$$ and $$B$$ have probabilities 0.25 and 0.50 respectively. The probability that both $$A$$ and $$B$$ occur simultaneously is 0.14. Then the probability that neither $$A$$ nor $$B$$ occurs is
The probability that an event $$A$$ happens in one trial of an experiment is 0.4. Three independent trials of the experiment are performed. The probability that the event $$A$$ happens at least once is
If $$A$$ and $$B$$ are two events such that $$P(A) > 0,$$ and $$P\left( B \right) \ne 1,$$ then $$P\left( {\frac{{\overline A }}{{\overline B }}} \right)$$ is equal to
(Here $$\overline A$$ and $$\overline B$$ are complements of $$A$$ and $$B$$ respectively).
A.
$$1 - P\left( {\frac{A}{B}} \right)$$
B.
$$1 - P\left( {\frac{{\overline A }}{B}} \right)$$
C.
$$\frac{{1 - P\left( {A \cup B} \right)}}{{P\left( {\overline B } \right)}}$$
D.
$$\frac{{P\left( {\overline A } \right)}}{{P\left( {\overline B } \right)}}$$