A few days ago
MATHS! A man throws three fair dice. Find the probability that all the three numbers are different.?
MATHS! A man throws three fair dice. Find the probability that all the three numbers are different.?
Top 3 Answers
A few days ago
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All the three numbers are different. If the first die is a one, then there are (5X4) 20 ways of getting the values on the other two dice different from the first one. Based on this there will 5 additional cases from two through six. Hence the total number of ways will be (6X5X4) = 120.
Based on the original sample space of 216 (6X6X6), the required probability will be 120/216 = [6X5X4]/216 = 5/9
Good luck in your studies,
~ Mitch ~
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4 years ago
what proportion factors on each and each die? Assuming 6, the 2nd die has 5/6 possibility of no longer being a similar variety because of the fact the 1st. If neither of the 1st 2 are a similar variety, the possibility the third would be diverse is 4/6. 5/6 * 4/6 = 20/36 = 5/9
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A few days ago
total no. of ways = 6 ^ 3 = 6 X 6 X 6 = 216
different faces can occur in 6! ways as different permutations of occurrence of number on face of dice
required probability = 6!
—-
216
= 120 / 216
= 0.555 or 0.56
i can guarantee u that this is right answer!!
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