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The probability is 11/36.

This is easiest to work out in the "negative":

Instead of working out all the "positive", or successful, outcomes and adding them together, the fact that the sum of all probabilities is 1 is used by working out all the "negative", or unsuccessful, outcomes, adding them together and subtracting this sum from 1:

For this case there are 3 positive (successful) outcomes: 6x, x6, 66 (where each pair represents the value showing on the dice (in order) with x meaning any digit 1-5) but only 1 negative (unsuccessful) outcome: xx.

Thus the calculation is:

Pr(at least one of the dice showing a six) = 1 - Pr(neither die shows a six).

The probability of a single die not showing a six is 5/6

→ the probability of two dice not showing a six is 5/6 × 5/6 = 25/36 as the two events are independent.

→ The probability of at least one six = 1 - probability of no six = 1 - 25/36 = 11/36.

It can be worked out in the positive, but requires much more work:

Pr(at least one 6) = Pr(6x: first die 6, second not 6) + Pr(x6: first not 6, second 6) + Pr (66: both 6)

= 1/6 × 5/6 + 5/6 × 1/6 + 1/6 × 1/6

= 5/36 + 5/36 + 1/36

= 11/36 (as before)

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Q: What are the odds when rolling 2 dice that at least one of them will be a 6?
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