Two different dice are rolled together find the probability of getting a doublet

And this problem We're rolling two dice and relate to all the probability At the some of the Dice Equals four. Now, how many ways can that happen? You might get a 202 or you might get a one and a three For three and a one. So remember in dice rolls we distinguish between the order here and if we do that, There's a total of 36 outcomes in the sample space, However, only three of them in the top have a sum that equals four. So three successes Over 36 total. Go ahead and simplify that.

Two different dice are thrown at the same time. Find the probability of getting :
(i) a doublet
(ii) a sum of 8

Solution:

When two dice are thrown simultaneously, the sample space of the experiment is

{(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)}

`1/6``1/2``3/4``5/6`

Solution : When two dice are thrown simultaneously, all possible outcomes are
(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),(4,5),(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).
Number of all possible outcomes = 36.
(i) Let `E_(1)` be the event of getting two numbers whose sum is 5.
Then, the favourable outcomes are (1,4),(2,3),(3,2),(4,1).
Number of favourable outcomes = 4.
`:.` P (getting two numbers whose sum is 5) = `P(E_(1)) = 4/36 = 1/9`.
(ii) Let `E_(2)` be the event of getting a even numbers on both dice.
Then, the favourable outcomes are
(2,2),(2,4),(2,6),(4,2),(4,4),(4,6),(6,2),(6,4),(6,6).
Number of favourable outcomes = 9.
`:. ` P(getting even number on both dice ) = `P(E_(2)) = 9/36 = 1/4`.
(iii) Let `E_(3)` be the event of getting a doublet.
Then, the favourable outcomes are
(1,2),(2,2),(3,3),(4,4),(5,5),(6,6).
Number of favourable outcomes = 6.
`:. ` P(getting a doublet ) = `P(E_(3)) = 6/36 = 1/6`.

Total number of possible outcomes = 36

(i) Doublets are (1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6 )

Total number of doublets = 6

∴ Probability (getting a doublet) = 6/36 or 1/6

(ii) Favourable outcomes are (4, 6), (5, 5), (6, 4) i.e., 3

∴ Probability (getting a sum 10) = 3/36 or 1/12

{(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)}

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