1211.

If P$\mathrm{P}\left(\mathrm{A} \cap \mathrm{B}\right) = \frac{7}{10} \mathrm{and} \mathrm{P}\left(\mathrm{B}\right) = \frac{17}{20}$, where P(A/B) stands for probability, then P(A/B) is equal to

• $\frac{7}{8}$

• $\frac{{\displaystyle 17}}{{\displaystyle 20}}$

• $\frac{14}{17}$

• $\frac{1}{8}$

A box has 100 pens of which 10 are defective. The probability that out of a sample of 5 pens drawn one by one with replacement and atmost one is defective, is

• $\frac{9}{10}$

• $\frac{1}{2}{\left(\frac{9}{10}\right)}^4$

• ${\left(\frac{9}{10}\right)}^5 + \frac{1}{2}{\left(\frac{9}{10}\right)}^4$

• $\frac{1}{2}{\left(\frac{9}{10}\right)}^5$

If ${}_{\mathrm{n}}\mathrm{C}_{12} = {}_{\mathrm{n}}\mathrm{C}_8$, then n is equal to

• 12

• 20

• 26

• 6

The probability distribution of X is

X	0	1	2	3
P(X)	0.3	k	2k	2k

The value of k is

• 0.7

• 0.3

• 1

• 0.14

Two events A and B will be independent if

• $\mathrm{P}\left(\mathrm{A}\text{'} \cap \mathrm{B}\text{'}\right) = \left(1 - \mathrm{P}\left(\mathrm{A}\right)\right)\left(1 - \mathrm{P}\left(\mathrm{B}\right)\right)$

• $\mathrm{P}\left(\mathrm{A}\right) + \mathrm{P}\left(\mathrm{B}\right) = 1$

• $\mathrm{P}\left(\mathrm{A}\right) = \mathrm{P}\left(\mathrm{B}\right)$

• A and B are mutually exclusive

There are four letters and four addressed envelopes. The chance that all letters are not despatched in the right envelope is

• 19/24

• 21/23

• 23/24

• None of these

The mean and variance ofa random variable X having a binomial distribution are 4 and 2 respectively, then P(X = 1) is

• $\frac{1}{32}$

• $\frac{1}{16}$

• $\frac{1}{8}$

• $\frac{1}{4}$

If the mean and standard deviation in probability binomial distribution are 3 and 3/2 respectively, then the probability distribution is

• ${\left(\frac{3}{4} + \frac{1}{4}\right)}^{12}$

• ${\left(\frac{1}{4} + \frac{3}{4}\right)}^{12}$

• ${\left(\frac{1}{4} + \frac{3}{4}\right)}^9$

• ${\left(\frac{3}{4} + \frac{1}{4}\right)}^9$

A student appears in the test I, II, III. Student is declared pass if he passes any two of three tests. The probability of the student of pass test I, II, III are p, q and $\frac{1}{2}$. If the probability of student's success is $\frac{1}{2}$, then

• p = 1, q = 0

• $\mathrm{p} = \frac{2}{3}, \mathrm{q} = \frac{1}{2}$

• p, q have infinitly many values

• All of the above

If A and B are the events for which $\mathrm{P}\left(\mathrm{A}\right) = \frac{1}{3}, \mathrm{P}\left(\mathrm{B}\right) = \frac{1}{4} \mathrm{and} \mathrm{P}\left(\mathrm{A} \cap \mathrm{B}\right) = \frac{1}{5}$, then $\mathrm{P}\left(\frac{\mathrm{A}}{\mathrm{B}}\right)$ is equal to

• $\frac{4}{5}$

• $\frac{3}{5}$

• $\frac{7}{5}$

• $\frac{1}{5}$