In a book of $$500$$ pages, it is found that there are $$250$$ typing errors. Assume that Poisson law holds for the number of errors per page. Then, the probability that a random sample of $$2$$ pages will contain no error, is :
A.
$${e^{ - 0.3}}$$
B.
$${e^{ - 0.5}}$$
C.
$${e^{ - 1}}$$
D.
$${e^{ - 2}}$$
Answer :
$${e^{ - 1}}$$
Solution :
Here number of errors per page $$ = \frac{{250}}{{500}} = \frac{1}{2}$$ and $$n = 2$$
$$\therefore \,\lambda = np = 2 \times \frac{1}{2} = 1$$
and probability of no error
$$P\left( {X = 0} \right) = \frac{{{e^{ - 1}} \times {{\left( 1 \right)}^0}}}{{0!}} = {e^{ - 1}}$$
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.
$$\frac{{P\left( {\overline A } \right)}}{{P\left( {\overline B } \right)}}$$