The person tells the truth 3 out of every 4 times.
Let's define:
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A : The die really shows six.
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B : The die does not show six.
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E : The person says the die shows six.
The chance of rolling a six when throwing a die,
P(A), is
‌.
The chance of not rolling a six,
P(B), is
‌.
If a six really comes up, the chance the person will say "six" (tells the truth) is
P(E∣A)=‌.
If a six did NOT come up, the chance the person wrongly says "six" (tells a lie) is
P(E∣B)=‌.
We want to find: What is the chance the die REALLY shows six if the person says it is a six? This is
P(A∣E).
Use Bayes' Theorem:
P(A∣E)=‌| P(A)⋅P(E∣A) |
| P(A)⋅P(E∣A)+P(B)⋅P(E∣B) |
Put in the values:
P(A∣E)=‌Solve:
P(A∣E)=‌=‌