Let A and B be events in a sample space S suchthat P(A) = 0.5, P(B) = 0.4 andP(A $\cup$ B) = 0.6. Observe the following lists

	List I		List II
(i)	$\mathrm{P}\left(\mathrm{A} \cap \mathrm{B}\right)$	(1)	0.4
(ii)	$\mathrm{P}\left(\mathrm{A} \cap \overline{\mathrm{B}}\right)$	(2)	0.2
(iii)	$\mathrm{P}\left(\overline{\mathrm{A}} \cap \mathrm{B}\right)$	(3)	0.3
(iv)	$\mathrm{P}\left(\overline{\mathrm{A}} \cap \overline{\mathrm{B}}\right)$	(4)	0.1

The correct match of List I from List II is

Two numbers are chosen at random from{1, 2, 3, 4, 5, 6, 7, 8} at a time. The probability that smaller of the two numbers is less than 4 is

• $\frac{7}{14}$

• $\frac{8}{14}$

• $\frac{9}{14}$

• $\frac{10}{14}$

263.

Two fair dice are rolled. The probability of the sum of digits on their faces to be greater than or equal to 10 is

• $\frac{1}{5}$

• $\frac{1}{4}$

• $\frac{1}{8}$

• $\frac{1}{6}$

A bag contains 2n + 1 corns. It is known that n of these coins have a head on both sides, whereas the remaining n + 1 coins are fair. A coin is picked up at random from the bag and tossed. If the probability that the toss results in a head is $\frac{31}{42}$, then n is equal to

• 10

• 11

• 12

• 13

The random variable takes the values 1, 2, 3, 1 ..., m. If P(X = n) = $\frac{1}{\mathrm{m}}$ to each n, then the variance of X is

• $\frac{\left(\mathrm{m} + 1\right)\left(2\mathrm{m} + 1\right)}{6}$

• $\frac{{\mathrm{m}}^2 - 1}{12}$

• $\frac{\mathrm{m} + 1}{2}$

• $\frac{{\mathrm{m}}^2 + 1}{12}$

$\mathrm{If} \mathrm{X} \mathrm{is} \mathrm{a} \mathrm{poisson} \mathrm{variate} \mathrm{P}\left(\mathrm{X} = 1\right) = 2\mathrm{P}\left(\mathrm{X} = 2\right), \mathrm{then} \mathrm{P}\left(\mathrm{X} = 3\right) = ?$

• $\frac{{\mathrm{e}}^{ - 1}}{6}$

• $\frac{\mathrm{e}{ }^{ - 2}}{2}$

• $\frac{{\mathrm{e}}^{ - 1}}{2}$

• $\frac{{\mathrm{e}}^{ - 1}}{3}$

The probability distribution of a random variable is given below

X = x	0	1	2	3	4	5	6	7
P(X = x)	0	k	2k	2k	3k	k2	2k2	7k2 + k

Then P(0 ) < X < 5) =?

• $\frac{1}{10}$

• $\frac{3}{10}$

• $\frac{8}{10}$

In a city, 10 accidents take place in a span of 50 days. Assuming that the number of accidents follow the Poisson distribution, the probability that three or more accidents occur in a day, is

• $\sum _{\mathrm{k} = 3}^{\infty } \frac{{\mathrm{e}}^{ - \mathrm{\lambda }}{\mathrm{\lambda }}^{\mathrm{k}}}{\mathrm{k}!}, \mathrm{\lambda } = 0.2$

• $\sum _{\mathrm{k} = 3}^{\infty } \frac{{\mathrm{e}}^{\mathrm{\lambda }}{\mathrm{\lambda }}^{\mathrm{k}}}{\mathrm{k}!}, \mathrm{\lambda } = 0.2$

• $1 - \sum _{\mathrm{k} = 0}^3 \frac{{\mathrm{e}}^{ - \mathrm{\lambda }}{\mathrm{\lambda }}^{\mathrm{k}}}{\mathrm{k}!}, \mathrm{\lambda } = 0.2$

• $\sum _{\mathrm{k} = 0}^3 \frac{{\mathrm{e}}^{ - \mathrm{\lambda }}{\mathrm{\lambda }}^{\mathrm{k}}}{\mathrm{k}!}, \mathrm{\lambda } = 0.2$

A pair of dice is thrown and sum of dice come up multiple of 4 then find probability that at least one dice shows 4

• $\frac{2}{7}$

• $\frac{4}{9}$

• $\frac{1}{9}$

• $\frac{5}{8}$

A bag contains 6 red and 10 green balls, 3 balls are drawn from it one by one without replacement. If the third ball drawn is red, then the probability, that first two balls are green is

• $\frac{3}{7}$

• $\frac{9}{149}$

• $\frac{9}{56}$

• $\frac{3}{8}$