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3 years ago

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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 tos

s. (a) What is the probability that we obtain a sum of 3 before we obtain a sum of 7

1 answer:

Simora [160]3 years ago

6 0

Answer:

$\frac{(2/36)}{(1-(28/36))} = 1/4$

Step-by-step explanation:

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