If the coefficients $a$ and $b$ of a quadratic expression $x^2+a x+b$ are chosen from the sets $A=\{3,4,5\}$ and $B=\{1,2,3,4\}$ respectively, then the probability that the equation $x^2+a x+b=0$ has real roots is

A random variable $X$ has the following probability distribution

$$ \begin{array}{|c|c|c|c|c|c|c|c|c|} \hline X=x & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 \\ \hline P(X=x) & 0.15 & 0.23 & K & 0.10 & 0.20 & 0.08 & 0.07 & 0.05 \\ \hline \end{array} $$

For the event $E=\{X / X$ is a prime number $\}$ and the event $F=\{X / X<4\}$, the probability $P(E \cup F)=$

4-digit numbers are formed using the digits 4, 5, 6, 7, 8, 9 allowing repetition of the given digits. If a number is chosen at random from those numbers thus formed, then the probability that it is exactly divisible by 3 is

If $E_1, E_2 \ldots, E_n$ are an independent events such that $P\left(E_r\right)=\frac{1}{1+r},(r=1,2, \ldots, n)$, then the probability that atleast one of $E_1, E_2, \ldots, E_n$ happens is