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

edited | 831 views
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Probability of getting total of 7 = (1/6)7.

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can u explain???
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We can get total of 7 only if we get 1 on each 7 dice.
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Sorry , I updated the question to 3 tosses , the answer given as 91 /216 . How is it possible ?
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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,

$P = \frac{1}{6^{7}}$
by Active (1.8k points)
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Sorry , I updated the question to 3 tosses , the answer given as 91 /216 . How is it possible ?
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What does atleast once mean here? Does (3, 4, 3) counts as 2 occurrences of total 7?

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
by Veteran (118k points)
edited by
+1

According to you answer should be 123.you add 6 instead of 3.

and you didn't subtract common combinations

(6,1,_) and (6,_,1)  common combination=6,1,1

(6,_,1) and (_,6,1)  common combination=6,6,1

(_,6,1) and (1,6,_)  common combination=1,6,1

(1,6,_) and (1,_,6)   common combination=1,6,6

(1,_,6) and (_,1,6)  common combination=1,1,6

(6,1,_)  and (_,1,6) common combination=6,1,6

same for  (5,2,_) , (4,3,_)

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yes rt. then total 9 possibilities will reduces
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only 9 ??it should be 18.Right?

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(6,1,_) can combine with  (6,_,1) , (_,1,6)

(6,_,1) now only combine with  (_,6,1) , no other possibility as (6,1,_) taken previously

(_,6,1)  with (1,6,_) ,(1,_,6)

(1,_,6) with (_,1,6)

yes 6 possibility each . tnks :)

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you did one more mistake in calculation you added 6 two times in total combination.you added 6 instead of 3.Please check.
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yes
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now I think answer is right.

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

by Boss (18.5k points)