Three students A, B and C can respectively solve 50%, 25% and 20% of the problems in a book. A particular problem is selected at random from the book. The probability that at least one of them will solve the problem 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 :

Bag A contains 3 red and 4 black balls, and Bag B contains 5 red and 'n' black balls. One ball is drawn at random from one of the bags and it is found to be red. If the probability that it was from bag B is 35/68, then the value of 'n' is:

Given that the events A and B are mutually exclusive such that $P(A)=\dfrac{1}{2}$, $P(A\cup B)=\dfrac{2}{3}$ and $P(B)=p$, then the value of $p$ is:

Let A and B be two events such that $P(A \cap B) = 0$ then $P(A|(A \cup B))$ is equal to:

A man is known to speak truth 3 out of 4 times. He throws a die and reports that it is four. The probability that it is actually four is

A pair of dice is thrown until the sum of numbers appeared is a perfect square or a non-perfect square sum appeared five times in succession. If random variable $X$ denotes the number of non perfect square sums appeared, then $P(X > 0)$ is