Question

In a simultaneous throw of a pair of dice, find the probability of getting -:
A double
a sum less then six
a sum more than 7
8 as sum

Answer 1

Step-by-step explanation:

1) a doublet

S={ (1,1) (2,2) (3,3) (4,4) (5,5) (6,6)}

P(doublet)=6/36=1/6

2) sum less than 6

S={(2,4)(3,3)(4,2)(5,1)} are the possibility we don't get... so,

n(s)=36.... but new n(s)= 36-4=32

P(of sum less than 6)= 32/36=16/18=8/9

3) sum mor than 7

S={ (2,6)(3,5)(3,6)(4,5)(4,6)(4,4)(5,3)(5,4)(5,6)(6,2)(6,4)(6,5)(6,6)} are the possibility we will get... so

P(sum >7)= 13/36

4) 8 as sum

S={(2,6)(3,5)(4,4)(5,3)(6,2)}

P(sum=8)= 5/36

Hope u got the answers sweetie.....☺️

Answer 2

Solution :-

Given : A simultaneous throw of a pair of dice.

Total Outcomes = 36

Formula used :

P(E) = F/T

i) A double

Favourable Outcomes = 6

P(getting a double) = 6/36 = 1/6

ii) A sum less than six

Favourable Outcomes = 10

P(getting a sum less than six) = 10/36 = 5/18

iii) A sum more than seven

Favourable Outcomes = 15

P( getting a sum more than seven) = 15/36

iv) 8 as sum

Favourable Outcomes = 5

P(getting 8 as sum) = 5/36