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24 Nov 2011, 11:22
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Can somebody help me understand this basic of probability please? "What is the probability of throwing 2 fair dices, getting first number always greater than the second number" . Please note, its not a sum from any book, but was used in one of the classes I attended as an explanation, which I could not still follow.
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24 Nov 2011, 11:51
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"What is the probability of throwing 2 fair dices, getting first number always greater than the second number"
For example
probability for the 1st dice is 1/6
probability for the 2nd dice is 1 - 1/6 = 5/6
So first number always greater than the second number
For example
probability for the 1st dice is 1/6
probability for the 2nd dice is 1 - 1/6 = 5/6
So first number always greater than the second number
24 Nov 2011, 13:18
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Here are the all possible outcomes.
(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)
All possible outcomes = 36
Favorable outcomes are those in green above = 15
Probability = Favorable outcomes/All possible outcomes = 15/36 = 5/12
(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)
All possible outcomes = 36
Favorable outcomes are those in green above = 15
Probability = Favorable outcomes/All possible outcomes = 15/36 = 5/12
