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:

Three companies $C_1, C_2, C_3$ produce car tyres. $A$ car manufacturing company buys $40 \%$ of its requirement from $C_1, 35 \%$ from $C_2$ and $25 \%$ from $C_3$. The company knows that $2 \%$ of the tyres supplied by $C_1, 3 \%$ by $C_2$ and $4 \%$ by $C_3$ are defective. If a tyre chosen at random from the consignment received is found defective, then the probability that it was supplied by $C_2$ is:

Recent studies suggest that $12\%$ of the world population is left-handed. Depending on parents' hand usage, the chances of having left-handed children are as follows: $A$. Both parents are left-handed, chances of having left-handed children = $24\%$. $B$. Both parents are right-handed, chance of having left-handed children = $9\%$. $C$. Father left-handed and mother right-handed, chances of having left-handed children = $17\%$. $D$. Father right-handed and mother left-handed, chances of having left-handed children = $22\%$. Given $P(A) = P(B) = P(C) = P(D) = 1/4$ and $L$ denotes the event that the child is left-handed. What is the probability $P(A|L)$?

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 ...... .

In a test,a student either guesses,copies,or knows the answer to a multiple-choice question with four choices. The probability that he guesses is $1/3$ and the probability that he copies the answer is $1/6$. The probability that his answer is correct,given that he copied it,is $1/8$. The probability that he knew the answer to the question,given that he answered it correctly,is

$A$ dealer gets refrigerators from $3$ different manufacturing companies $C_1, C_2$ and $C_3$. $25 \%$ of his stock is from $C_1, 35 \%$ from $C_2$ and $40 \%$ from $C_3$. The percentages of receiving defective refrigerators from $C_1, C_2$ and $C_3$ are $3 \%, 2 \%$ and $1 \%$ respectively. If a refrigerator sold at random is found to be defective by a customer, then the probability that it is from $C_2$ is