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1 answer:

3 0

There are the combinations that result in a total less than 7 and at least one die showing a 3:

[3, 3] [3,2] [2,1] [1,3] [2,3]
The probability of each of these is 1/6 * 1/6 = 1/36

There is a little ambiguity here about whether or not we should count [3,3] as the problem says "and one die shows a 3." Does this mean that only one die shows a 3 or at least one die shows a 3? Assuming the latter, the total probability is the sum of the individual probabilities:

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

Therefore, the required probability is: 5/36

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