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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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thaovtp1407

Answer:

Step-by-step explanation:

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

P ([tex]w_{2} =2[/tex]) =[tex]\frac{1}{6} *\frac{1}{6}[/tex] = [tex]\frac{1}{36}[/tex] ≠ [tex]\frac{1}{11}[/tex]

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