A six faced die is biased such that

$$3 \times \mathrm{P}($$a prime number$$)\,=6 \times \mathrm{P}($$a composite number$$)\,=2 \times \mathrm{P}(1)$$.

Let X be a random variable that counts the number of times one gets a perfect square on some throws of this die. If the die is thrown twice, then the mean of X is :

Let $$S$$ be the sample space of all five digit numbers. It $$p$$ is the probability that a randomly selected number from $$S$$, is a multiple of 7 but not divisible by 5 , then $$9 p$$ is equal to :

Let $$X$$ be a binomially distributed random variable with mean 4 and variance $$\frac{4}{3}$$. Then, $$54 \,P(X \leq 2)$$ is equal to :

The mean and variance of a binomial distribution are $$\alpha$$ and $$\frac{\alpha}{3}$$ respectively. If $$\mathrm{P}(X=1)=\frac{4}{243}$$, then $$\mathrm{P}(X=4$$ or 5$$)$$ is equal to :