Bob is asked to construct a probability model for rolling a pair of fair dice. He lists the outcomes as​ 2, 3,​ 4, 5,​ 6, 7,​ 8, 9,​ 10, 11,

Question

Bob is asked to construct a probability model for rolling a pair of fair dice. He lists the outcomes as​ 2, 3,​ 4, 5,​ 6, 7,​ 8, 9,​ 10, 11, 12. Because there are 11​ outcomes, he​ reasoned, the probability of rolling a nine must be one eleventh . What is wrong with​ Bob’s reasoning

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Raelynn 2 hours 2021-10-14T08:25:57+00:00 1 Answer 0

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    2021-10-14T08:27:23+00:00

    Answer:

    Step-by-step explanation:

    Bob is wrong because he rolled only 11 time. Let call w_{2} is the outcome with the sum of pints in two die are 2. Then :

    P (w_{2} =2) =\frac{1}{6} *\frac{1}{6} = \frac{1}{36}\frac{1}{11}

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45:7+7-4:2-5:5*4+35:2 =? ( )