Question

The person with all correct get's brainliest ..​

Answer 1

Step-by-step explanation:

number 3=1/4

number 4=2/5

number 5=1/3

Answer 2

$\displaystyle\huge\red{\underline{\underline{Solution}}}$

3.

The dice is rolled twice

So the total number of possible outcomes =

${6}^{2} = 36$

Let A be the event that OF GETTING PRIME NUMBER ON EACH DICE

Then the points are

$= \:(2,2),(2,3),(2,5),(3,2),(3,3),(3,5),(5,2),(5,3),(5,5)$

So the total number of possible outcomes for the event A is = 9

So the required probability = P(A)

$= \frac{9}{36} = \frac{1}{4}$

4.

The dice is rolled twice

The dice is rolled twice So the total number of possible outcomes =

${6}^{2} = 36$

Let B be the event that OF GETTING MULTIPLE OF 2 IN FIRST DICE AND MULTIPLE OF 3 IN SECOND DICE

Then the points are

$= (2,3)(2,6),(4,3),(4,6),(6,3),(6,6)$

So the total number of possible outcomes for the event B is =

is = So the required probability = P(B)

$= \frac{6}{36} = \frac{1}{6}$

5.

The dice is rolled twice

The dice is rolled twice So the total number of possible outcomes =

${6}^{2} = 36$

Let D be the event that OF GETTING SUM OF BOTH NUMBERS LESS THAN 8

Then the points are

(1,1),(1,2),(1,3),(1,4),(1,5),(1,6),(2,1),(2,2),(2,3),(2,4),(2,5),(3,1),(3,2),(3,3),(3,4),(4,1),(4,2),(4,3),(5,1),(5,2),(6,1)

So the total number of possible outcomes for the event D is = 21

So the required probability = P(D) =

$\frac{21}{36} = \frac{7}{12}$