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17 Nov 2018, 10:05
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When 7 fair standard 6-sided dice are thrown, the probability that the sum of the numbers on the top faces is 10 can be written as
\(\frac{n}{6^7}\), where n is a positive integer. What is n?
A) 42
B) 49
C) 56
D) 63
E) 84
17 Nov 2018, 11:26
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For sum to be 10
Three 2s & Four 1s
Number of ways= 7!/(4!x3!)= 35
One 4 & Six 1s
Number of ways= 7!/(6!)= 7
Thus total ways= 42
Hence answer= 42/6^7
Three 2s & Four 1s
Number of ways= 7!/(4!x3!)= 35
One 4 & Six 1s
Number of ways= 7!/(6!)= 7
Thus total ways= 42
Hence answer= 42/6^7
17 Nov 2018, 11:42
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The minimum on one face of a die is 1. If all die show 1, then the sum will only be 7. Also, thinking about the other way, if one of the die shows 6, then even if others have minimum value of 1, sum is greater than 10, This is also true for any one die showing 5. So, maximum value can be 4. So possible cases are:
4, 1, 1, 1, 1, 1, 1
This can be arranged in \(\frac{7!}{6!}\) ways = 7
3, 2, 1, 1, 1, 1, 1
This can be arranged in \(\frac{7!}{5!}\) ways = 7*6 = 42
2, 2, 2, 1, 1, 1, 1
This can be arranged in \(\frac{7!}{(4!*3!)}\) ways = \(\frac{(7*6*5)}{(3*2)}\) = 35
Total = 7 + 42 + 35 = 84, hence n = 84
Option E
4, 1, 1, 1, 1, 1, 1
This can be arranged in \(\frac{7!}{6!}\) ways = 7
3, 2, 1, 1, 1, 1, 1
This can be arranged in \(\frac{7!}{5!}\) ways = 7*6 = 42
2, 2, 2, 1, 1, 1, 1
This can be arranged in \(\frac{7!}{(4!*3!)}\) ways = \(\frac{(7*6*5)}{(3*2)}\) = 35
Total = 7 + 42 + 35 = 84, hence n = 84
Option E
