Two players play the following game: $A$ writes $3,5,6$ on three different cards: $B$ writes $8,9,10$ on three different cards. Both draw randomly two cards from their collections. Then, $A$ computes the product of two numbers helshe has drawn, and $B$ computes the sum of two numbers he/she has drawn. The player getting the larger number wins. What is the probability that A wins?
$\frac{1}{3}$
$\frac{5}{9}$
$\frac{4}{9}$
$\frac{1}{9}$
An integer is chosen at random and squared. The probability that the last digit of the square is $1$ or $5$ is
The corners of regular tetrahedrons are numbered $1, 2, 3, 4.$ Three tetrahedrons are tossed. The probability that the sum of upward corners will be $5$ is
Two dice are thrown. The events $A, B$ and $C$ are as follows:
$A:$ getting an even number on the first die.
$B:$ getting an odd number on the first die.
$C:$ getting the sum of the numbers on the dice $\leq 5$
Describe the events $A^{\prime }.$
Two integers $\mathrm{x}$ and $\mathrm{y}$ are chosen with replacement from the set $\{0,1,2,3, \ldots ., 10\}$. Then the probability that $|x-y|>5$ is:
The chance of throwing a total of $7$ or $12$ with $2$ dice, is