$A, B_1, B_2, B_3$ are the events in a random experiment. If $P(B_1)=0.25, P(B_2)=0.30, P(B_3)=0.45, P(A|B_1)=0.05, P(A|B_2)=0.04, P(A|B_3)=0.03$, then $P(B_2|A) = $

There are $3$ bags which are known to contain $2$ white and $3$ black balls; $4$ white and $1$ black balls and $3$ white and $7$ black balls respectively. $A$ ball is drawn at random from one of the bags and found to be a black ball. Then the probability that it was drawn from the bag containing the most black balls is

$A$ man is known to speak the truth $2$ out of $3$ times. If he throws a die and reports that it is $6$,then the probability that it is actually $5$ is:

There are $3$ bags,each containing $5$ white balls and $3$ black balls. Also,there are $2$ bags,each containing $2$ white balls and $4$ black balls. $A$ white ball is drawn at random. Find the probability that this white ball is from a bag of the first group.

$A$ doctor assumes that a patient has one of three diseases $d_1, d_2,$ or $d_3$. Before any test,he assumes an equal probability for each disease. He carries out a test that will be positive with probability $0.7$ if the patient has disease $d_1$,$0.5$ if the patient has disease $d_2$,and $0.8$ if the patient has disease $d_3$. Given that the outcome of the test was positive,what is the probability that the patient has disease $d_2$?

$A$ boy speaks the truth in $3$ out of $5$ times. If he throws a die and reports that the number appeared on it is $5$, then the probability that it is actually $5$ is: