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sweet [91]
3 years ago
15

Hey guys, Do you guys know the answer to this TTM problem? Thank You!

Mathematics
1 answer:
tensa zangetsu [6.8K]3 years ago
4 0
I think it is 2.5,2.5
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Suppose you are given either a fair dice or an unfair dice (6-sided). You have no basis for considering either dice more likely
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Answer: Our required probability is 0.83.

Step-by-step explanation:

Since we have given that

Number of dices = 2

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Probability of getting a fair dice P(E₁) = \dfrac{1}{2}

Number of unfair dice = 1

Probability of getting a unfair dice  P(E₂) = \dfrac{1}{2}

Probability of getting a 3 for the fair dice P(A|E₁)= \dfrac{1}{6}

Probability of getting a 3 for the unfair dice P(A|E₂) = \dfrac{1}{3}

So, we need to find the probability that the die he rolled is fair given that the outcome is 3.

So, we will use "Bayes theorem":

P(E_1|A)=\dfrac{P(E_1)P(A|E_1)}{P(E_1)P(A|E_1)+P(E_2)P(A|E_2)}\\\\(E_1|A)=\dfrac{0.5\times 0.16}{0.5\times 0.16+0.5\times 0.34}\\\\P(E_1|A)=0.83

Hence, our required probability is 0.83.

8 0
3 years ago
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