Show Me A 6!

Number of cases when all the dice show a number other than $6$ = $5\times 5\times 5$ = $125$ times. Total possible outcomes= $6\times 6\times 6$ = $216$ times. Thus number of outcomes in which at least one die shows a " $6$ "= $216-125=91$ Therefore required probability is $\frac { 91 }{ 216 }$ .

A new approach

Solve it by using binomial distribution .

The probability of getting 6 on one throw is 1/6. The dice is thrown thrice . So the required probability is the 3C1(5/6)^2×(1/6)+3C2(5/6)×(1/6)^2+3C3(1/6)^3= 91/216

Case 1: Exactly 1 six

1x5x5=25, since the dice are distinct (shown in the picture), we multiply 25 by 3, thus, 75 events

Case 2: Exactly 2 six

1x1x5=5, again, we multiply 5 by 3, thus, 15 events

Case 3: Exactly 3 six

There is only one event for this case.

Now, the probability of having at least one six show up is 75+15+1 divided by all events (216).

Thus, the answer is 91/216.

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Total number of outcomes is 6 X 6 X 6 =216.

Total number of possibilities to get the at least one six is =5 X 5 X 5 =125,

so we have {1-(125/216) }=91/216 is the required answer.

probability of getting 1 six= 1-probability of getting no six that is 1-5/6 5/6 5/6=91/226