SOLUTION: Sally claims to have telekinetic powers. She claims that she can cause a die to turn up any number she chooses. A fair die is produced, and Sally predicts that she will roll a six

Question 1186715: Sally claims to have telekinetic powers. She claims that she can cause a die to turn up any number she chooses.
A fair die is produced, and Sally predicts that she will roll a six 3 times in a row. She proceeds to do just that. Our evidence and hypothesis are as follows:
E: Sally rolls a six 3 times in a row (without cheating).
H: Sally has telekinetic powers.
Given that Sally has telekinetic powers, E is just the sort of thing we should expect. So, assume that P(E/H) = 1.
But prior to considering E, we are understandably skeptical, giving odds of 1 to 9 that Sally has telekinetic powers.
Assuming chance is the only alternative to Sally’s alleged telekinetic powers, what is P(E/∼H)? (Hint: You will need to use the restricted conjunction rule.)
What is P(H/E)?
I have managed to solve the problem. But I am not sure if it's correct because I don't think I used the restricted conjunction rule, which is P(A·B) = P(A) x P(B)
Here's my solution, please tell me if am correct or wrong:
* P(E) = 1/6^3 = 1/216
* P(H) = 1/10
* P(~H) = 9/10
* P(E|H) = 1
what is P(E/∼H)?
* using conditional rule:
P(E|~H) = P(~H·E) / P(~H)
= [ P(~H) * P(E) ] / P(~H)
= P(E)
= 1/216
What is P(H/E)?
* using Bayes’ theorem:
P(H|E) = [ P(H)*P(E|H) ] / [ P(H)*P(E|H) + P(~H)*P(E/~H) ]
= [ (1/10)*(1) ] / [ (1/10)*(1) + (9/10)*(1/216) ]
= 0.96

Answer by CPhill(1959) (Show Source): You can put this solution on YOUR website!
Here's how to break down this Bayesian reasoning problem:
**1. Probability of E given not H (P(E|¬H)):**
If Sally *doesn't* have telekinetic powers, the probability of rolling a six on a fair die is 1/6. Since the rolls are independent, the probability of rolling a six three times in a row is:
P(E|¬H) = (1/6) * (1/6) * (1/6) = (1/6)³ = 1/216 ≈ 0.0046
**2. Prior probabilities:**
The odds of Sally having telekinetic powers are 1 to 9. This means:
P(H) / P(¬H) = 1/9
Since P(H) + P(¬H) = 1, we can solve for P(H) and P(¬H):
P(H) = 1/10 = 0.1
P(¬H) = 9/10 = 0.9
**3. Applying Bayes' Theorem:**
We want to find P(H|E), the probability that Sally has telekinetic powers *given* that she rolled three sixes in a row. Bayes' Theorem states:
P(H|E) = [P(E|H) * P(H)] / [P(E|H) * P(H) + P(E|¬H) * P(¬H)]
**4. Plugging in the values:**
P(H|E) = [1 * 0.1] / [1 * 0.1 + (1/216) * 0.9]
P(H|E) = 0.1 / [0.1 + 0.004167]
P(H|E) = 0.1 / 0.104167
P(H|E) ≈ 0.960
**Therefore:**
* P(E|¬H) ≈ 0.0046
* P(H|E) ≈ 0.960

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