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Probability (Discussion)
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Sureshbala said – Thu, 16 Oct 2008 05:55:29 -0000 ( Link )
Since we need at least one six, let us calculate the negation of this…I mean calculate the probability of not getting a six on any of these three dice.
The probability of not getting a six on a die is 5/6
Since the three events here are independent events, the probability of not getting six on any of these three dice is 5/6×5/6×5/6 = 125/216.
Hence the probability of getting at least one six is 1-125/216 = 91/216.
In fact you can also solve this directly by calculating what we need.
At least one six gives us the following cases
Case1: Exactly one six Case 2: Exactly two sixes Case 3: Exactly three sixes
Considering Case 1: The probability of getting exactly one six is 1/6×5/6×5/6 and since one six could be from any of the three dice in
ways, the probability is 3(1/6×5/6×5/6) = 75/216
Considering Case 2: The probability of getting exactly two sixes is 1/6×5/6×5/6 ans sine two sixes could be from any of the three dice in
ways, the probability is 3(1/6×1/6×5/6) = 15/216
Considering Case 3: The probability of getting six on all the three dice is 1/6×1/6×1/6 = 1/216.
Hence the probability of getting at least one six is 75/216+15/216+1/216 = 91/216.
But this will take some time compared to what we have in the first case.
I hope this is clear
Regards, Suresh
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ankita said – Thu, 16 Oct 2008 07:40:49 -0000 ( Link )
Hi Sureshbala, That’s right ! Thanks for the explanation.
I would like to share alternate way :
Find the number of cases in which none of the digits show a ‘6’.
i.e. all three dice show a number other than ‘6’, 5 * 5 *5 = 125 cases.
Total possible outcomes when three dice are thrown = 216.
The number of outcomes in which at least one die shows a ‘6’ = Total possible outcomes when three dice are thrown – Number of outcomes in which none of them show ‘6’.
= 216 – 125 = 91.
The required probability = 91 / 216
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