if you answer this i will make you as brainliest. two dice are numbered 1,2,3,4,5,6 and 1,1,2,2,3,3 respectivly.they are rolled and the sum of the numbers on them is noted.find the probability of getting each sum from 2 to 9 seperately.
Answers
Favourable outcomes 36
(1,1) (2,1) (3,1) (4,1) (5,1) (6,1)
(1,1) (2,1) (3,1) (4,1) (5,1) (6,1)
(1,2) (2,2) (3,2) (4,2) (5,2) (6,2)
(1,2) (2,2) (3,2) (4,2) (5,2) (6,2)
(1,3) (2,3) (3,3) (4,3) (5,3) (6,3)
(1,3) (2,3) (3,3) (4,3) (5,3) (6,3)
Sum of 2=1/18
Sum of 3=1/9
Sum of 4=1/9
Sum of 5=1/6
Sum of 6=1/6
Sum of 7=2/9
Sum of 8=1/9
Sum of 9=1/18
Step-by-step explanation:
Total number of dice thrown = 2.
Possible outcomes n(S) = 6² = 36.
(i) Sum as 2.
Let E be the event of getting the sum of 2.
E = (1,1), (1,1)
n(E) = 2
∴ Required probability P(E) = n(E)/n(S) = 2/36 = 1/18.
(ii) Sum as 3:
Let E₁ be the event of getting the sum as 3.
E₁ = (1,2),(2,1),(1,2), (2,1)
n(E₁) = 4
∴ Required probability P(E₁) = n(E₁)/n(S) = 1/9.
(iii) Sum as 4:
Let E₂ be the event that the sum is 4.
E₂ = (3,1),(1,3),(2,2),(2,2)
n(E₂) = 4
∴ Required probability P(E₂) = n(E₂)/n(S) = 1/9
(iv) Sum as 5:
Let E₃ be the event of getting sum as 5.
E₃ = (4,1),(1,4),(2,3),(3,2), (3,2),(2,3)
n(E₃) = 6.
Required probability P(E₃) = 6/36 = 1/6
(v) Sum as 6:
n(E₄) = (1,5),(5,1),(2,4),(4,2),(3,3),(3,3) = 6
Required probability = 6/36 = 1/6
(vi) Sum as 7:
n(E₅) = (6,1),(1,6),(4,3),(4,3),(4,3),(4,3),(5,2),(5,2) = 8
∴ Required probability = 8/36 = 2/9.
(vii) Sum as 8:
n(E₆) = (6,2),(6,2),(5,3),(5,3) = 4.
∴ Required probability = 4/36 = 1/9
(vii) Sum as 9:
n(E₇) = (6,3),(6,3) = 2
∴ Required probability = 2/36 = 1/18.
Hope it helps!