Let S be the sample space of the random experiment of throwing simultaneously two unbiased dice with six faces (numbered1 to 6) and let E_k = {(a, b) ∈ S : ab = k} for k $\ge$ 1. If p_k + P(E_k) for k $\ge$ 1, then the correct among the following, is

• ${\mathrm{p}}_1 < {\mathrm{P}}_{30} < {\mathrm{P}}_4 < {\mathrm{P}}_6$

• ${\mathrm{p}}_{36} < {\mathrm{P}}_6 < {\mathrm{P}}_2 < {\mathrm{P}}_4$

• ${\mathrm{p}}_1 < {\mathrm{P}}_{11} < {\mathrm{P}}_4 < {\mathrm{P}}_6$

• ${\mathrm{p}}_{36} < {\mathrm{P}}_{11} < {\mathrm{P}}_6 < {\mathrm{P}}_4$

For k = 1, 2, 3 the box B_k contains k red balls and        (k + 1) white balls. Let P(B₁) = $\frac{1}{2}$, P(B₂) = $\frac{1}{3}$ and P(B₃) = $\frac{1}{6}$. A box is selected at 36 random and a ball is drawn from it. If a redball is drawn, then the probability that it has come from box B, is

• $\frac{35}{78}$

• $\frac{14}{39}$

• $\frac{10}{13}$

• $\frac{12}{13}$

$\mathrm{The} \mathrm{distribution} \mathrm{of} \mathrm{a} \mathrm{random} \mathrm{variable} \mathrm{X} \mathrm{is} \mathrm{given} \mathrm{below}$

X = x	- 2	- 1	0	1	2	3
P(X = x)	1/10	k	1/5	2k	3/10	k

• $\frac{1}{10}$

• $\frac{2}{10}$

• $\frac{3}{10}$

• $\frac{7}{10}$

254.

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

• $\frac{1}{2{\mathrm{e}}^2}$

• $\frac{1}{3{\mathrm{e}}^2}$

• $\frac{2}{3{\mathrm{e}}^2}$

• $\frac{1}{{\mathrm{e}}^2}$

$$\begin{gathered} \mathrm{A} \mathrm{and} \mathrm{B} \mathrm{are} \mathrm{events} \mathrm{of} \mathrm{a} \mathrm{random} \mathrm{experiment} \\ \mathrm{such} \mathrm{that} \mathrm{P}\left(\mathrm{A} \mathrm{U} \mathrm{B}\right) = \frac{4}{5}, \mathrm{P}\left(\overline{\mathrm{A}} \mathrm{U} \overline{\mathrm{B}}\right) = \frac{7}{10} \mathrm{and} \mathrm{P}\left(\mathrm{B}\right) = \frac{2}{5}, \mathrm{then} \mathrm{P}\left(\mathrm{A}\right) = ? \end{gathered}$$

• $\frac{9}{10}$

• $\frac{8}{10}$

• $\frac{7}{10}$

• $\frac{3}{5}$

Suppose that E₁ and E₂ are two events of a random experiment such that P(E₁) = 1/4, P(E₂/E₁) and P(E₁/E₂) = 1/4, observe the lists given below

        List I                        List II

(A)    P(E₂)                  (i) 1/4

(B)    $\mathrm{P}\left({\mathrm{E}}_1 \cup {\mathrm{E}}_2\right)$           (ii) 5/8

(C)   $\mathrm{P}\left(\overline{{\mathrm{E}}_1}/ \overline{{\mathrm{E}}_2}\right)$             (iii) 1/8

(D)   $\mathrm{P}\left({\mathrm{E}}_1/\overline{{\mathrm{E}}_2}\right)$              (iv) 3/8

                                  (v) 3/8

                                  (vi) 3/4

The correct matching of the List I from the List II is

If A_i (i = 1, 2, 3, ... , n) are n independent events with P(A_i) = $\frac{1}{1 + \mathrm{i}}$ for each i, then the probability that none of A_i occurs is

• $\frac{\mathrm{n} - 1}{\mathrm{n} + 1}$

• $\frac{\mathrm{n}}{\mathrm{n} + 1}$

• $\frac{\mathrm{n}}{\mathrm{n} + 2}$

• $\frac{1}{\mathrm{n} + 1}$

Suppose A and B are two events such that $\mathrm{P}\left(\mathrm{A} \cap \mathrm{B}\right) = \frac{3}{25} \mathrm{and} \mathrm{P}\left(\mathrm{B} - \mathrm{A}\right) = \frac{8}{25}. \mathrm{Then}, \mathrm{P}\left(\mathrm{B}\right)$ is equal to

• $\frac{11}{25}$

• $\frac{3}{11}$

• $\frac{1}{11}$

• $\frac{9}{11}$

Suppose that a random variable X follows Poisson distribution. If P(X = 1) = P(X = 2) then P(X = 5) is equal to

• $\frac{2}{3}{\mathrm{e}}^{- 2}$

• $\frac{3}{4}{\mathrm{e}}^{- 2}$

• $\frac{4}{15}{\mathrm{e}}^{- 2}$

• $\frac{7}{8}{\mathrm{e}}^{- 2}$

If the mean and variance of a binomial variable X are 2 and 1 respectively, then P(X $\ge$ 1) is equal to

• $\frac{2}{3}$

• $\frac{15}{16}$

• $\frac{7}{8}$

• $\frac{4}{5}$