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Data Interview Question

Is It Raining in Seattle?

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Solution & Explanation

Problem Restatement

You have three friends in Seattle, each with a 2/3 chance of telling the truth and a 1/3 chance of lying. All three friends tell you it is raining. You need to determine the probability that it is actually raining.

Frequentist Approach

  1. Understanding the Problem:

    • Each friend independently tells the truth with a probability of 2/3.
    • All three friends say it is raining.
    • We want to calculate the probability that it is indeed raining.
  2. Calculate the Probability of All Friends Lying:

    • If all friends are lying, then it is not raining.

    • Probability that a single friend lies: 1/3.

    • Probability that all three friends lie:

      P(All Lie)=(13)3=127P(\text{All Lie}) = \left(\frac{1}{3}\right)^3 = \frac{1}{27}

  3. Calculate the Probability of Raining:

    • The probability it is not raining is the probability that all friends are lying.

    • Thus, the probability it is raining is:

      P(Rain)=1P(All Lie)=1127=2627P(\text{Rain}) = 1 - P(\text{All Lie}) = 1 - \frac{1}{27} = \frac{26}{27}

Bayesian Approach

  1. Assumptions:

    • Let's assume the prior probability of rain in Seattle is 50% (though this is not given in the problem, it's a common assumption for such problems).
    • We define FiF_i as the event that friend ii says it is raining.
  2. Calculate Conditional Probabilities:

    • Probability all friends say it is raining given it is raining (all tell the truth):

      P(F1F2F3R)=(23)3=827P(F_1 \cap F_2 \cap F_3 | R) = \left(\frac{2}{3}\right)^3 = \frac{8}{27}

    • Probability all friends say it is raining given it is not raining (all lie):

      P(F1F2F3Rc)=(13)3=127P(F_1 \cap F_2 \cap F_3 | R^c) = \left(\frac{1}{3}\right)^3 = \frac{1}{27}

  3. Apply Bayes' Theorem:

    • We use Bayes' theorem to find the probability it is raining given all friends say it is raining:

      P(RF1F2F3)=P(F1F2F3R)P(R)P(F1F2F3)P(R | F_1 \cap F_2 \cap F_3) = \frac{P(F_1 \cap F_2 \cap F_3 | R) \cdot P(R)}{P(F_1 \cap F_2 \cap F_3)}

  4. Calculate Total Probability:

    • Total probability that all friends say it is raining:

      P(F1F2F3)=P(F1F2F3R)P(R)+P(F1F2F3Rc)P(Rc)P(F_1 \cap F_2 \cap F_3) = P(F_1 \cap F_2 \cap F_3 | R) \cdot P(R) + P(F_1 \cap F_2 \cap F_3 | R^c) \cdot P(R^c)

      =8270.5+1270.5=854+154=954=16= \frac{8}{27} \cdot 0.5 + \frac{1}{27} \cdot 0.5 = \frac{8}{54} + \frac{1}{54} = \frac{9}{54} = \frac{1}{6}

  5. Final Calculation:

    • Substitute back into Bayes' theorem:

      P(RF1F2F3)=8270.516=89P(R | F_1 \cap F_2 \cap F_3) = \frac{\frac{8}{27} \cdot 0.5}{\frac{1}{6}} = \frac{8}{9}

Conclusion

The Bayesian approach yields a probability of 89\frac{8}{9} that it is raining given all friends say it is raining, under the assumption of a 50% prior probability of rain. The frequentist approach, focusing on the probability that at least one friend is truthful, gives a probability of 2627\frac{26}{27}. The difference arises due to the assumptions and focus of each method.