Probability

$\mathrm{If} \frac{1 + 3\mathrm{P}}{3}, \frac{1 - 2\mathrm{P}}{2} \mathrm{are} \mathrm{probabilities} \mathrm{of} \mathrm{two} \mathrm{mutually} \mathrm{exclusive} \mathrm{events}, \mathrm{then} \mathrm{P} \mathrm{lies} \mathrm{in} \mathrm{the} \mathrm{interval}$

• $\left[- \frac{1}{3}, \frac{1}{2}\right]$

• $\left(\frac{- 1}{2}, \frac{1}{2}\right)$

• $\left[- \frac{1}{3}, \frac{2}{3}\right]$

• $\left(\frac{- 1}{3}, \frac{2}{3}\right)$

A speaks truth in 75% of the cases and B in 80% of the cases. Then, the probability that their statements about an incident do not match, is

• $\frac{7}{20}$

• $\frac{3}{20}$

• $\frac{2}{7}$

• $\frac{5}{7}$

$\mathrm{Given} \mathrm{below} \mathrm{is} \mathrm{the} \mathrm{distribution} \mathrm{of} \mathrm{a} \mathrm{random} \mathrm{variable} \mathrm{X}$

If $\mathrm{\alpha } = \mathrm{P}\left(\mathrm{X} < 3\right) \mathrm{and} \mathrm{\beta } = \mathrm{P}\left(\mathrm{X} > 2\right), \mathrm{then} \mathrm{\alpha } : \mathrm{\beta } = ?$

• 2 5

• 3 4

• 4 5

• 3 7

${\mathrm{sech}}^{ - 1}\left(\frac{1}{\sqrt{2}}\right) + {\mathrm{csch}}^{ - 1}\left( - 1\right) = ?$

• 0

• $\sqrt{2} + 1$

• $\sqrt{2}$

• $\sqrt{2} - 1$

The number of different ways of preparing a garland using 6 distinct white roses and 5 distinct red roses such that no two red roses come together is

• 21600

• 43200

• 86400

• 151200

Box I contains 30 cards numbered I to 30 and Box II contains 20 cards numbered 31 to 50. A box is selected at random and a card is drawn from it. The number on the card is found to be a non-prime number. The probability that the card was drawn from Box I is :

• $\frac{2}{3}$

• $\frac{2}{5}$

• $\frac{8}{17}$

• $\frac{4}{17}$

Let E^c denote the complement of an event E. Let E₁, E₂ and E₃ be any pairwise independent events withP(E₁) > 0 and $\mathrm{P}\left({\mathrm{E}}_1 \cap {\mathrm{E}}_2\cap {\mathrm{E}}_3\right) = 0 \mathrm{then} \left({{\mathrm{E}}_3}^{\mathrm{c}} \cap \raisebox{1ex}{${{\mathrm{E}}_3}^{\mathrm{c}}$}\!\left/ \!\raisebox{-1ex}{${\mathrm{E}}_1$}\right.\right)$ is equal to

• $\mathrm{P}\left({{\mathrm{E}}_3}^{\mathrm{c}}\right) - \mathrm{P}\left({\mathrm{E}}_2\right)$

• $\mathrm{P}\left({{\mathrm{E}}_3}^{\mathrm{c}}\right) - \mathrm{P}\left({{\mathrm{E}}_2}^{\mathrm{c}}\right)$

• $\mathrm{P}\left({\mathrm{E}}_3\right) - \mathrm{P}\left({\mathrm{E}}^{\mathrm{c}}\right)$

• $\mathrm{P}\left({{\mathrm{E}}_2}^{\mathrm{C}}\right) + \mathrm{P}\left({\mathrm{E}}_3\right)$

Two families with three members each and one family with four members are to be seated in a row. In how many ways can they be seated so that the same family members are not separated ?

• 2! 3! 4!

• 3!(4!)³

• 3! . 2 . (4!)

• (3!)³ . (4!)