If the mean and variance of a binomial distribution are 4 and $\frac{4}{3}$ respectively, then $P(X=2)=$

If a man throws a die until he gets a number bigger than 3 , then the probability that he gets a 5 in his last throw is

A diagnostic test has the probability 0.95 of giving a positive result when applied to a person suffering from a certain disease and a probability 0.10 of giving a positive result when given to a non-sufferer. It is estimated that $0.5 \%$ of the population are suffering from the disease. If this test is now administered to a person from this population about whom there is no information relating to the incidence of this disease and the test gives a positive result, then the probability that he is a sufferer, is

Consider the following statements

Assertion (A) If $P_1, P_2, P_3$ are probability of happening of three independent events, then probability of happening of atleast one of them is $1-\left[\left(1-P_1\right)\left(1-P_2\right)\left(1-P_3\right)\right]$

Reason (R) For any three independent events $A, B$ and $C$

$$ \begin{array}{r} P(A \cup B \cup C)=P(A)+P(B)+P(C)-P(A) P(B)-P(A) P(C) -P(B) P(C)+P(A) P(B) P(C) \end{array} $$

The correct option among the following is