Match List - I with List - II.

List - I	List - II
(A) Two events E and F will be independent if $P(E'F')$ is equal to	(I) $1 - P(E/F)$
(B) If $P(F) \neq 0$, then $P(E'/F)$ is equal to	(II) $P(E) = P(F)$
(C) If E and F are independent events, then	(III) $P(E \cap F') = P(E) \cdot P(F')$
(D) If $P(E \cap F) \neq 0$ and $P(E/F) = P(F/E)$, then	(IV) $[1-P(E)][1-P(F)]$

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A problem in mathematics is given to three students whose chances of solving it are 1/2, 1/3, 1/4 respectively. The probability that the problem is solved is

The probability of a man hitting a target is 1/2. How many times must he fire so that the probability of hitting the target at least once is more than 90%?

Match List-I with List-II

Let $A$ and $B$ be two events such that $P(A) = 0.2$, $P(B) = 0.4$, $P(B|A) = 0.5$

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A beg contains 4 white and 6 red balls. Four balls are drawn at random, then match the probability of the numbers of white balls drawn. Where P(E) denotes the probability of an event E.

Then Match List-I with List-II

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A random variable X has the following probability distribution

Then P (X < 3) is

If we take 8 identical slips of paper and write the number 0 on one of them, the number 1 on three of the slips, the number 2 on three of the slips and the number 3 on one of the slips. These slips are folded, put in a box and roughly mixed. One slip is drawn at random from the box. If X is the random variable denoting the number written on the drawn slip, the variance of X is: