161.

The probability that atleast one of the events A and B occurs is 0.6. If A and B occurs simultaneously with probability 0.2,then $\mathrm{P}\left(\overline{\mathrm{A}}\right) + \mathrm{P}\left(\overline{\mathrm{B}}\right)$ is

• 0.4

• 0.8

• 1.2

• 1.4

If X is a poisson variate such that P (X = 1) = P (X = 2), then P (X = 4) is equal to

• $\frac{1}{2{\mathrm{e}}^2}$

• $\frac{1}{3{\mathrm{e}}^2}$

• $\frac{2}{3{\mathrm{e}}^2}$

• $\frac{1}{{\mathrm{e}}^2}$

The chances of defective screws in three boxes A, B, C are $\frac{1}{5}, \frac{1}{6}, \frac{1}{7}$ respectively. A box 5 6 7 is selected at random and a screw drawn from it at random is found to be defective. Then, the probability that it came from box A, is

• $\frac{16}{29}$

• $\frac{1}{15}$

• $\frac{27}{59}$

• $\frac{42}{107}$

The value of $\frac{\cos \left(\mathrm{\theta }\right)}{1 + \sin \left(\mathrm{\theta }\right)}$ is equal to

• $\tan \left(\frac{\mathrm{\theta }}{2} - \frac{\mathrm{\pi }}{4}\right)$

• $\tan \left(- \frac{\mathrm{\pi }}{4} - \frac{\mathrm{\theta }}{2}\right)$

• $\tan \left(\frac{\mathrm{\pi }}{4} - \frac{\mathrm{\theta }}{2}\right)$

• $\tan \left(\frac{\mathrm{\pi }}{4} + \frac{\mathrm{\theta }}{2}\right)$

The probability that a certain kind of component will survive a given shock test is $\frac{3}{4}$· The probability that exactly 2 of the next 4 components tested survive is

• $\frac{9}{41}$

• $\frac{25}{128}$

• $\frac{1}{5}$

• $\frac{27}{128}$

A random variable X follows binomial distribution with mean $\mathrm{\alpha }$ and variance $\mathrm{\beta }$ Then

• $0 < \mathrm{\alpha } < \mathrm{\beta }$

• $0 < \mathrm{\beta } < \mathrm{\alpha }$

• $\mathrm{\alpha } < 0 < \mathrm{\beta }$

• $\mathrm{\beta } < 0 < \mathrm{\alpha }$

If the probability density function of a random variable X is f(x) = $\frac{\mathrm{x}}{2}$ in $0 \le \mathrm{x} \le 2$, then $\mathrm{P}\left(\mathrm{X} > 1.5 | \mathrm{x} > 1\right)$ is equal to :

• $\frac{7}{16}$

• $\frac{3}{4}$

• $\frac{7}{12}$

• $\frac{21}{64}$

If X follows a binomial distribution with parameters n = 100 and p = $\frac{1}{3}$ then P(X = r) 3 is maximum when r is equal to

• 16

• 32

• 33

• none of these

A random variable X takes values 0, 1, 2, 3, ... with probability P(X = x) = k(x + 1)${\left(\frac{1}{5}\right)}^{\mathrm{x}}$, where k is constant, then P(X = 0) is

• 7/25

• 18/25

• 13/25

• 16/25

Out of 15 persons 10 can speak Hindi and 8 can speak English. If two persons are chosen at random, then the probability that one person speaks Hindi only and the other speaks both Hindi and English is

• 3/5

• 7/12

• 1/5

• 2/5