$A$ person who tosses an unbiased coin gains two points for turning up a head and loses one point for a tail. If three coins are tossed and the total score $X$ is observed, then the range of $X$ is

$A$ random variable $X$ takes the values $0, 1, 2$. Its mean is $1.2$. If $P(X=0)=0.3$,then $P(X=1)=$

$A$ fair die is tossed repeatedly until a six is obtained. Let $X$ denote the number of tosses required and let $a=P(X=3)$,$b=P(X \geq 3)$ and $c=P(X \geq 6 \mid X>3)$. Then $\frac{b+c}{a}$ is equal to

India plays two matches each with West Indies and Australia. In any match,the probabilities of India getting $0, 1,$ and $2$ points are $0.45, 0.05,$ and $0.50$ respectively. Assuming that the outcomes are independent,the probability of India getting at least $7$ points is:

$A$ die is loaded in such a way that each odd number is twice as likely to occur as each even number. If $E$ is the event that a number greater than or equal to $4$ occurs on a single toss of the die,then $P(E)$ is equal to:

$A$ random variable $X$ has the following probability distribution:
$X = 1, 2, 3, 4, 5$
$P(X) = 0.1, 0.2, 0.3, 0.2, 0.2$
For the events $E = \{X \text{ is a prime number}\}$ and $F = \{X < 4\}$, find $P(E \cup F)$.