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What is the probability of getting total of 7  atleast  once in 3 tosses of a fair dice... ?
 

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In a toss of 7 fair dices, a total of 7 is possible only if all tosses result in 1 as outcome, therefore it makes only one possibility out of all possible i.e., $6^{7}$ outcomes, Therefore,

$P = \frac{1}{6^{7}}$
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getting atleast 7 in 3 tosses

(6,1,_) , (5,2,_) , (4,3,_), (1,1,5) , (1,2,4) , (1,3,3) , (1,4,2) , (2,2,3)

Now combination of (6,1,_)=6

                              (6,_,1)=6

                              (_,6,1)=6

                              (1,6,_)=6

                               (1,_,6)=6

                               (_,1,6)=6

Such combination also possible for (5,2) and (4,3)

Now for (1,1,5) = 3 combination

(1,2,4)=6 combination

(2,2,3)=3 combination

(1,3,3)=3 combination

Total combination 36+36+36+3+3+6+3 - 18(for repeated elements)=105

So, probability = 105/216
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Total sample space = 63

Uisng method of bars and stars:   0 0 |  0 0 0 | 0 0

Now, there are 2 options:

1) We can get 7 in any 2 tosses and you dont bother of the number on the other toss.

    Lets say, you complete total of 7 in first 2 tosses and you dont bother of the number on the last toss.

    so, x1 + x2 = 7 (x1>0 and x2>0)(for first 2 tosses)     which gives (2+5-1)C(5) = 6C5 = 6

   We dont care of the number on the last dice.

   So, total = 6*6 = 36. This is one combination.

   You can have 3C2 combinations.

   SO, total = 3C2 * 36 = 108

2)  we get total of 7 comprising all the dices.

    x1 + x2 +x3 = 7 (x1>0 and x2>0 and x3>0) which gives (3+4-1)C(4) = 6C4 = 15

So, probablity = (108 + 15)/216 = 123/216

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