Question

two dice. are thrown find the probability of getting a prime no. less than 8​

Answer 1

IF DICE ARE ROLLED WE GET 36 OUTCOMES

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

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

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

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

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

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

n{S} = 36

WHEN WE ADD NUMBERS IN BRACKET WE GET FOLLOWING OUTCOMES

S = (2,3,4,5,6,7,

3,4,5,6,7,8,

4,5,6,7,8,9,

5,6,7,8,9,10

6,7,8,9,10,11

7,8,9,10,11,12)

SUPPOSE 'A' IS THE EVENT THAT SHOWS PRIME NUMBER LESS THAN 8 IN ABOVE

A = (2,3,5,7)

n(A) = 4

SO..

P(A) = n(A) /n(S)

P(A) = 4/36

P(A) = 1/9

WHEN WE MULTIPLY IT WITH EACH OTHER IN BRACKET WE GET FOLLOWING OUTCOMES

S = (1,2,3,4,5,6,

2,4,6,8,10,12,

3,6,9,12,15,18,

4,8,12,16,20,24,

5,10,15,20,25,30

6,12,18,24,30,36)

SUPPOSE 'B' THE EVENT THAT SHOWS THE PRIME NUMBER LESS THAN 8 IN ABOVE

B = (2,3,5)

n(B) = 3

SO..

P(B) = n(B) /n(S)

P(B) = 3/36

P(B) = 1/12

I HOPE THIS HELPFUL TO YOU