Question

A pair of six face dice with faces marked from 1 to 6 thoroughly simultaneously.what is the probability that sum of numbers which shows up

Answer 1

Step-by-step explanation:

Given, a pair of dices thrown once

⇒ total no. of outcomes =36

E: Product of two number appearing on dice is to be divisible by 5.

No. of Outcomes=(1,5),(5,1),(2,5),(5,2),(3,5),(5,3),(4,5),(5,4),(5,5),(5,6),(6,5)

P(E)= 11/36

F: Atleast one of dice shows the face 1.

No. of Outcomes=(1,1),(1,2),(2,1),(1,3),(3,1),(1,4),(4,1),(1,5),(5,1),(1,6),(6,1)

P(F)= 11/36

and P(E∩F)= 2/36 = 1/18

since P(E∩F)≠ 0

⇒ E,F are not mutually exclusive and P(E∩F) is not equal to P(E).P(F).

⇒ E,F are not independent.

Since P(E)=P(F)⇒ E,F are equiprobable too.

$\huge{ son \: prodigal}$