Bag $I$ contains $3$ red and $2$ green balls and Bag $II$ contains $5$ red and $3$ green balls. $A$ ball is drawn from one of the bags at random and it is found to be green. Then the probability that it is drawn from Bag $I$ is

There are two dice $A$ and $B$. Die $A$ has $4$ red and $2$ white faces,and die $B$ has $2$ red and $4$ white faces. $A$ coin is tossed once. If it shows head,die $A$ is rolled; if it shows tail,die $B$ is rolled. If the probability that die $A$ is used is $\left(\frac{32}{33}\right)$ given that red turns up every time in the first $n$ throws,then $n=$

Box $I$ contains $30$ cards numbered $1$ to $30$ and Box $II$ contains $20$ cards numbered $31$ to $50$. $A$ box is selected at random and a card is drawn from it. The number on the card is found to be a non-prime number. The probability that the card was drawn from Box $I$ is

$A$ bag contains $2n+1$ coins. It is known that $n$ of these coins have a head on both sides, whereas the remaining $n+1$ coins are fair. $A$ coin is picked up at random from the bag and tossed. If the probability that the toss results in a head is $\frac{31}{42}$, then $n$ is equal to

$A$ company produces $10,000$ items per day. On a particular day, $2500$ items were produced on machine $A$, $3500$ on machine $B$, and $4000$ on machine $C$. The probability that an item produced by machines $A, B, C$ is defective is $2 \%$, $3 \%$, and $5 \%$, respectively. If one item is selected at random from the output and is found to be defective, then the probability that it was produced by machine $C$ 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: