The probability that a non-leap year selected at random will have $53$ Sundays is

Given $P(A \cup B)=0.6$ and $P(A \cap B)=0.2$,the probability that exactly one of the events occurs is:

The probabilities of three students solving a question are $\frac{1}{2}, \frac{1}{4},$ and $\frac{1}{6}$ respectively. What is the probability that the question is solved?

$A$ number is chosen at random from the first ten natural numbers. The probability that the number is odd and a perfect square is

For independent events $A_1, A_2, \dots, A_n$,$P(A_i) = \frac{1}{i + 1}$ for $i = 1, 2, \dots, n$. Then the probability that none of the events will occur is:

Let $S$ be the sample space of the random experiment of throwing simultaneously two unbiased dice with six faces (numbered $1$ to $6$) and let $E_k = \{(a, b) \in S : ab = k\}$ for $k \geq 1$. If $p_k = P(E_k)$ for $k \geq 1$,then which of the following is correct?