$A$ box $A$ contains $2$ white,$3$ red and $2$ black balls. Another box $B$ contains $4$ white,$2$ red and $3$ black balls. If two balls are drawn at random,without replacement,from a randomly selected box and one ball turns out to be white while the other ball turns out to be red,then the probability that both balls are drawn from box $B$ is

Two balls are selected at random one by one without replacement from a bag containing $4$ white and $6$ black balls. If the probability that the first selected ball is black,given that the second selected ball is also black,is $\frac{m}{n}$,where $\operatorname{gcd}(m, n) = 1$,then $m + n$ is equal to :

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

$A$ 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

$A$ 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

$A$ bag contains $6$ balls. Two balls are drawn from it at random and both are found to be black. The probability that the bag contains at least $5$ black balls is