For any two events A and B, which of the following are correct?

(A) $P(A\cup B) = P(A)+P(B)-P(A\cap B)$

(B) $P(A) = P(A\cup B)+P(A\cap B)$

(C) $P(A/B) = P(A)/P(B)$

(D) $P(A) = P(A\cap B)+P(A\cap \bar{B})$

Choose the correct answer from the options given below:

If A and B are two independent events such that $P(A) = 0.4$, and $P(B) = 0.5$, then P (neither A nor B) is

Let $X$ denote the number of hours you play during a randomly selected day. The probability that $X$ can take values $x$ has the following form, where $c$ is some constant.

$\mathrm{P}(\mathrm{X}=\mathrm{x})=\left\{\begin{array}{lll} 0.1, & \text { if } \mathrm{x}=0 \\ \mathrm{cx}, & \text { if } \mathrm{x}=1 \text { or } \mathrm{x}=2 \\ \mathrm{c}(5-\mathrm{x}), & \text { if } \mathrm{x}=3 \text { or } \mathrm{x}=4 \\ 0, & \text { otherwise } \end{array}\right.$

Match List-I with List-II :

A card is picked at random from a pack of 52 playing cards. If the picked card is a queen, then probability of card to be of spade type also, is

If the probability of two successes is 9 times the probability of 3 successes in 3 trials of a binomial distribution, then the probability of success in each trial is:

There are $6$ cards numbered $1$ to $6$, one number on one card. Two cards are drawn at random without replacement. Let X denote the sum of the numbers on the two cards drawn. Then $\mathrm{P}(\mathrm{X}>3)$ is :

Bag I contains 3 red and 4 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 red. The probability that it was drawn from Bag II is :