Three screws are drawn at random from a lot of $50$ screws containing $5$ defective ones. The probability of the event that all $3$ screws drawn are non-defective,assuming that the drawing is $(a)$ with replacement and $(b)$ without replacement,is respectively:

If $12$ identical balls are to be placed randomly in $3$ identical boxes,then the probability that one of the boxes contains exactly $3$ balls is

For three events $A, B$ and $C$,$P(\text{Exactly one of } A \text{ or } B \text{ occurs}) = P(\text{Exactly one of } B \text{ or } C \text{ occurs}) = P(\text{Exactly one of } C \text{ or } A \text{ occurs}) = \frac{1}{4}$ and $P(\text{All the three events occur simultaneously}) = \frac{1}{16}$. Then the probability that at least one of the events occurs is:

$A$ box contains $100$ balls,numbered from $1$ to $100$. If $3$ balls are selected one after the other at random with replacement from the box,then the probability that the sum of the three numbers on the balls selected is an odd number,is

Let $B_{i} (i=1, 2, 3)$ be three independent events in a sample space. The probability that only $B_{1}$ occurs is $\alpha$,only $B_{2}$ occurs is $\beta$,and only $B_{3}$ occurs is $\gamma$. Let $p$ be the probability that none of the events $B_{i}$ occurs,and these $4$ probabilities satisfy the equations $(\alpha - 2\beta)p = \alpha\beta$ and $(\beta - 3\gamma)p = 2\beta\gamma$ (All the probabilities are assumed to lie in the interval $(0, 1)$). Then $\frac{P(B_{1})}{P(B_{3})}$ is equal to ..........

If a year is selected at random from the $22^{nd}$ century,what is the probability that it contains $53$ Sundays (in $/28$)?