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Suppose that two balanced, six sided dice are tossed repeatedly and the sum of the two uppermost faces is determined on each toss. (a) What is the probability that we obtain a sum of 3 before we obtain a sum of 7

Sagot :

Answer:

[tex]\frac{(2/36)}{(1-(28/36))} = 1/4[/tex]

Step-by-step explanation:

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