Two dice are thrown independently. Let $$\mathrm{A}$$ be the event that the number appeared on the $$1^{\text {st }}$$ die is less than the number appeared on the $$2^{\text {nd }}$$ die, $$\mathrm{B}$$ be the event that the number appeared on the $$1^{\text {st }}$$ die is even and that on the second die is odd, and $$\mathrm{C}$$ be the event that the number appeared on the $$1^{\text {st }}$$ die is odd and that on the $$2^{\text {nd }}$$ is even. Then :

In a binomial distribution $$B(n,p)$$, the sum and the product of the mean and the variance are 5 and 6 respectively, then $$6(n+p-q)$$ is equal to :

A bag contains 6 balls. Two balls are drawn from it at random and both are found to be black. The probability that the bag contains at least 5 black balls is :

If an unbiased die, marked with $$-2,-1,0,1,2,3$$ on its faces, is thrown five times, then the probability that the product of the outcomes is positive, is :