A person tossing a biased coin indefinitely wins the game by getting head for the first time. The probability that he wins the game in odd number of tosses is $3 / 4$. If 5 such coins are tossed at a time then the probability that head appears on all the coins is

Let $B(\alpha, \beta, \gamma)$ represents that a bag $B$ contains $\alpha$ red balls, $\beta$ green balls and $\gamma$ blue balls. Given $B_1(2,3,2), B_2(3,2,2), B_3(2,2,3)$. A die is rolled. If the die shows up 2 or 3 or 5 , then a ball will be drawn at random from bag $B_1$. If the die shows up 4 or 6 , then a ball will be drawn at random from bag $B_2$. If the die shows up 1 , then from bag $B_3$ a ball will be drawn at random. Then the probability of drawing a green ball from a bag thus chosen is

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 $P(E \cup F)=$