In the random experiment of tossing two unbiased dice let E be the event of getting the sum 8 and F be the event of getting even numbers on both the dice. Then :

I. P(E) = $\frac{7}{36}$ II. p (F) = $\frac{1}{3}$

Which of the following is a correct statement ?

• Both I and II are true

• Neither I nor II is true

• I is true, II is false

• I is false, II is true

The probability distribution of a random variable X is given by

The variance of X is

• 1.76

• 2.45

• 3.2

• 4.8

$$\begin{gathered} \mathrm{If} \mathrm{A} \mathrm{and} \mathrm{B} \mathrm{are} \mathrm{independent} \mathrm{events} \mathrm{of} \mathrm{arandom} \mathrm{experiment} \mathrm{such} \mathrm{that} \\ \mathrm{P}(\mathrm{A} \cap \mathrm{B}) = \frac{1}{6} \mathrm{and} \mathrm{P}\left(\overline{\mathrm{A}} \cap \overline{\mathrm{B}}\right) = \frac{1}{3}, \mathrm{then} \mathrm{P}\left(\mathrm{A}\right) = ? \end{gathered}$$

• $\frac{1}{4}$

• $\frac{1}{3}$

• $\frac{1}{2}$

• $\frac{2}{3}$

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

• $\frac{35}{78}$

• $\frac{14}{39}$

• $\frac{10}{13}$

• $\frac{12}{13}$

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

• $\frac{1}{10}$

• $\frac{2}{10}$

• $\frac{3}{10}$

• $\frac{7}{10}$

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

• $\frac{1}{2{\mathrm{e}}^2}$

• $\frac{1}{3{\mathrm{e}}^2}$

• $\frac{2}{3{\mathrm{e}}^2}$

• $\frac{1}{{\mathrm{e}}^2}$

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

• $\frac{9}{10}$

• $\frac{8}{10}$

• $\frac{7}{10}$

• $\frac{3}{5}$

159.

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

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

• $\frac{\mathrm{n}}{\mathrm{n} + 1}$

• $\frac{\mathrm{n}}{\mathrm{n} + 2}$

• $\frac{1}{\mathrm{n} + 1}$