Bag $1$ contains $4$ white balls and $5$ black balls,and Bag $2$ contains $n$ white balls and $3$ black balls. One ball is drawn randomly from Bag $1$ and transferred to Bag $2$. $A$ ball is then drawn randomly from Bag $2$. If the probability that the ball drawn from Bag $2$ is white is $29/45$,then $n$ is equal to:

In an entrance test, there are multiple-choice questions. There are four possible answers to each question, of which one is correct. The probability that a student knows the answer to a question is $9/10$. If he gets the correct answer to a question, then the probability that he was guessing is

$A$ box $B_1$ contains $3$ blue balls and $6$ red balls. Another box $B_2$ contains $8$ blue balls and $n$ red balls $(n \in N)$. $A$ ball selected at random from a box is found to be red. If $p$ is the probability that this red ball drawn is from box $B_2$,then

$A$ 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 $d_2$?

$A$ bag contains $8$ balls,whose colors are either white or black. $4$ balls are drawn at random without replacement and it was found that $2$ balls are white and $2$ balls are black. The probability that the bag originally contained an equal number of white and black balls is:

In a group of $400$ people,$160$ are smokers and non-vegetarian,$100$ are smokers and vegetarian,and the remaining $140$ are non-smokers and vegetarian. Their chances of getting a particular chest disorder are $35\, \%, 20 \,\%$,and $10 \,\%$ respectively. $A$ person is chosen from the group at random and is found to be suffering from the chest disorder. The probability that the selected person is a smoker and non-vegetarian is ...... .