The probability that $A$ speaks the truth is $\frac{4}{5}$. $A$ coin is tossed. $A$ reports that a head appears. The probability that there was actually a head is

For $k=1, 2, 3$,the box $B_k$ contains $k$ red balls and $(k+1)$ white balls. Let $P(B_1) = \frac{1}{2}$,$P(B_2) = \frac{1}{3}$,and $P(B_3) = \frac{1}{6}$. $A$ box is selected at random and a ball is drawn from it. If a red ball is drawn,then the probability that it comes from box $B_2$ is:

Four boxes $A, B, C$ and $D$ contain $5000, 3000, 2000$ and $1000$ fuses respectively. The percentages of defective fuses in these boxes are $3\%, 2\%, 1\%$ and $0.5\%$ respectively. If a fuse selected at random from one of the boxes is found to be defective, then the probability that it has come from box $D$ is

$A$ shopkeeper buys a particular type of electric bulbs from three manufacturers $M_1, M_2$,and $M_3$. He buys $25 \%$ of his requirement from $M_1$,$45 \%$ from $M_2$,and $30 \%$ from $M_3$. Based on past experience,he found that $2 \%$ of type $M_3$ bulbs are defective,whereas only $1 \%$ of type $M_1$ and type $M_2$ bulbs are defective. If a bulb chosen by him at random is defective,then the probability that it was of type $M_3$ is

Two coins are available, one fair and the other two-headed. Choose a coin and toss it once; assume that the unbiased coin is chosen with probability $\frac{3}{4}$. Given that the outcome is head, the probability that the two-headed coin was chosen, is

Three boxes $B_1$,$B_2$ and $B_3$ contain balls with different colors as follows:

Box	White,Black,Red
$B_1$	$2, 1, 2$
$B_2$	$3, 2, 4$
$B_3$	$4, 3, 2$

$A$ die is thrown. Box $B_1$ is chosen if either $1$ or $2$ turns up. Box $B_2$ is chosen if $3$ or $4$ turns up and box $B_3$ is chosen if $5$ or $6$ turns up. Having chosen a box in this way,a ball is drawn at random from that box. If the ball drawn is found to be Red,then the probability that it is drawn from box $B_2$ is