Five numbers $${x_1},{x_2},{x_3},{x_4},{x_5}$$ are randomly selected from the numbers 1, 2, 3, ......., 18 and are arranged in the increasing order $$({x_1} < {x_2} < {x_3} < {x_4} < {x_5})$$. The probability that $${x_2} = 7$$ and $${x_4} = 11$$ is :

Let a biased coin be tossed 5 times. If the probability of getting 4 heads is equal to the probability of getting 5 heads, then the probability of getting atmost two heads is :

A biased die is marked with numbers 2, 4, 8, 16, 32, 32 on its faces and the probability of getting a face with mark n is $${1 \over n}$$. If the die is thrown thrice, then the probability, that the sum of the numbers obtained is 48, is :

Let E₁ and E₂ be two events such that the conditional probabilities $$P({E_1}|{E_2}) = {1 \over 2}$$, $$P({E_2}|{E_1}) = {3 \over 4}$$ and $$P({E_1} \cap {E_2}) = {1 \over 8}$$. Then :