Hi
the probability of getting 4,5, or 6 on the first toss
and a 1,2,3 or 4 on the second toss.
P = 12/36 = 1/3
(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)
Wish You the Best in your Studies.
.
The probability of getting 4, 5, or 6 on the first toss is = .
The probability of getting 1, 2, 3, or 4 on the second toss is = .
The outcomes of the first and the second toss are INDEPENDENT, so the final answer is the product
P = = . ANSWER
Solved.
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To solve the problem, there is NO any need to present the table of the 36 outcomes, as @ewatrrr does.
There is no any need to spend your time to construct and to write this table,
as well as there is no any need to spend your time counting the number of favorable outcomes from this table.