For a biased die,the probabilities for different faces to turn up are

$Face$	$1$	$2$	$3$	$4$	$5$	$6$
$P(F)$	$0.2$	$0.22$	$0.11$	$0.25$	$0.05$	$0.17$

The die is tossed and you are told that either face $4$ or face $5$ has turned up. The probability that it is face $4$ is

$A$ cubical die with faces marked $1, 2, 3, ..., 6$ is tossed such that the probability of throwing the number $t$ is proportional to $t^2$. The probability that the number $5$ has appeared,given that the number turned up is not even,is:

Let $B(\alpha, \beta, \gamma)$ represent 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 $2, 3,$ or $5$, a ball is drawn from bag $B_1$. If the die shows $4$ or $6$, a ball is drawn from bag $B_2$. If the die shows $1$, a ball is drawn from bag $B_3$. Find the probability of drawing a green ball.

Bag $P$ contains $3$ white, $2$ red, $5$ blue balls and bag $Q$ contains $2$ white, $3$ red, $5$ blue balls. $A$ ball is chosen at random from $P$ and is placed in $Q$. If a ball is chosen from bag $Q$ at random, then the probability that it is a red ball is

If $E_1$ and $E_2$ are two events of a random experiment such that $P(E_1) = \frac{1}{8}$,$P(E_1 \mid E_2) = \frac{1}{3}$,and $P(E_2 \mid E_1) = \frac{1}{4}$,then match the items of List-$I$ with the items of List-$II$.

List-$I$	List-$II$
$A. P(E_1 \cup E_2)$	$I. \frac{3}{29}$
$B. P(E_2)$	$II. \frac{26}{29}$
$C. P(E_1 \mid \bar{E}_2)$	$III. \frac{3}{16}$
$D. P(\bar{E}_1 \mid \bar{E}_2)$	$IV. \frac{3}{32}$

Find $P(E | F)$ for a coin tossed three times,where $E: \text{head on third toss}$,$F: \text{heads on first two tosses}$.