The probability that in the random arrangement of the letters of the word 'UNIVERSITY', the two I’s does not come together is :
A.
$$\frac{4}{5}$$
B.
$$\frac{1}{5}$$
C.
$$\frac{1}{{10}}$$
D.
$$\frac{9}{{10}}$$
Answer :
$$\frac{4}{5}$$
Solution :
Total number of arrangements of the letters of the word UNIVERSITY is $$\frac{{10!}}{{2!}}.$$
Number of arrangements when both I's are together $$ = 9!$$
So. the number of ways in which $$2$$ I’s do not together $$\frac{{10!}}{{2!}} - 9!$$
$$\therefore $$ Required probability
$$\eqalign{
& = \frac{{\frac{{10!}}{{2!}} - 9!}}{{\frac{{10!}}{{2!}}}} \cr
& = \frac{{10! - 9!\,2!}}{{10!}} \cr
& = \frac{{10! \times 9! - 9!.2!}}{{10!}} \cr
& = \frac{{9!\left[ {10 - 2} \right]}}{{10 \times 9!}} \cr
& = \frac{8}{{10}} \cr
& = \frac{4}{5} \cr} $$
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.
$$\frac{{P\left( {\overline A } \right)}}{{P\left( {\overline B } \right)}}$$