Question

please solve fast........​

Answer 1

Answer:

solution: When two dice are thrown simultaneously, thus number of event can be 62 = 36 because each die has 1 to 6 number on its faces. ...

Answer 2

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

Two dices are thrown simultaneously

So the total number of possible outcomes

$= {6}^{2} = 36$

1.

Let A be the event that the SUM OF BOTH FACES IS EVEN NUMBER

Then the event points for the event A are

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

So the total number of possible outcomes for the event A is 18

So the required probability is

$\displaystyle \: P(A) = \frac{18}{36} = \frac{1}{2}$

2.

Let B be the event that the RESULT IS A DOUBLET

Then the event points for the event B are

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

So the total number of possible outcomes for the event B is 6

So the required probability is

$\displaystyle \: P(B) = \frac{ 6}{36} = \frac{1}{6}$

3.

Let D be the event that the SUM OF BOTH FACES IS 9

Then the event points for the event D are

(3,6),(4,5),(5,4),(6,3)

So the total number of possible outcomes for the event D is 4

So the required probability is

$\displaystyle \: P(D) = \frac{4}{36} = \frac{1}{9}$

4.

Let E be the event that the SUM OF BOTH FACES IS LESS THAN 5

Then the event points for the event E is

(1,1),(1,2),(2,1),(1,3),(2,2),(3,1)

So the total number of possible outcomes for the event E is 6

So the required probability is

$\displaystyle \: P(E) = \frac{ 6}{36} = \frac{1}{6}$