If an unbiased coin is tossed 10 times, probability of obtaining more head than tail is :

It is given that only 0.1% of a large population have COVID infection. In this population, the reliability of COVID RTPCR-test is specified as follows :

For persons having COVID, 90% of the test detects the disease but 10% goes undetected. For persons not having COVID, 99% of the test is judged COVID negative but 1% are diagnosed as COVID positive.

Based on the above informations, answer the question :

The probability that randomly selected person from a population, not having COVID is :

For a Binomial distribution, B(n,p), where p+q=1, the sum and product of mean and variance are 8 and 12 respectively, when the value of n is:

If $P(A) = \frac{3}{10}$, $P(B) = \frac{2}{5}$ and $P(A \cup B) = \frac{3}{5}$ then the value of $P(B|A) + P(A|B)$ is:

A random variable X has the following probability distribution:

The value of $P(4 < x < 7)$ is equal to

Bag I contains 4 red and 5 black balls, while another Bag II contains 5 red and 6 black balls. One ball is drawn at random from one of the bags and it is found to be black. Then the probability that it was drawn from Bag II, is

If A is any event associated with sample space and If $E_1, E_2, E_3$ are mutually exclusive and exhaustive events. Then which of the following are true?

(A) $P(A) = P(E_1)P(E_2|A) + P(E_2)P(E_2|A) + P(E_3)P(E_3|A)$

(B) $P(A) = P(A|E_1)P(E_1) + P(A|E_2)P(E_2) + P(A|E_3)P(E_3)$

(C) $P(E_i|A) = \frac{P(A|E_i)P(E_i)}{\sum_{i=1}^3 P(A|E_i)P(E_i)}$, $i = 1,2,3$

(D) $P(A|E_i) = \frac{P(E_i|A)P(E_i)}{\sum_{i=1}^3 P(E_i|A)P(E_i)}$, $i = 1,2,3$

Choose the correct answer from the options given below: