If $4\,P(A) = 6\,P(B) = 10\,P(A \cap B) = 1,$ then $P\left( \frac{B}{A} \right) = $

One of the two events must occur. If the chance of one is $\frac{2}{3}$ of the other,then the odds in favour of the other are

If $P(A) = 0.65$ and $P(B) = 0.15$,then $P(\bar{A}) + P(\bar{B}) = $

Five horses are in a race. $Mr. A$ selects two of the horses at random and bets on them. The probability that $Mr. A$ selected the winning horse is (in $/5$)

One ticket is selected at random from $100$ tickets numbered $00, 01, 02, \dots, 98, 99$. If $X$ and $Y$ denote the sum and the product of the digits on the tickets respectively,then $P(X = 9 | Y = 0)$ equals:

$A$ card is drawn from a well-shuffled deck of $52$ cards. Find the probability of drawing a face card.