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

• $\frac{1}{10}$

• $\frac{2}{10}$

• $\frac{3}{10}$

• $\frac{7}{10}$

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

• $\frac{1}{2{\mathrm{e}}^2}$

• $\frac{1}{3{\mathrm{e}}^2}$

• $\frac{2}{3{\mathrm{e}}^2}$

• $\frac{1}{{\mathrm{e}}^2}$

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

• $\frac{9}{10}$

• $\frac{8}{10}$

• $\frac{7}{10}$

• $\frac{3}{5}$

256.

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

 

A. (A) (B) (C) (D)	(i) (ii) (iii) (vi) (i)
B. (A) (B) (C) (D)	(ii) (iv) (v) (vi) (i)
C. (A) (B) (C) (D)	(iii) (iv) (ii) (vi) (i)
D. (A) (B) (C) (D)	(iv) (i) (ii) (iii) (iv)

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

• $\frac{2}{3}$

• $\frac{15}{16}$

• $\frac{7}{8}$

• $\frac{4}{5}$