Calculate the conditional probability of a die showing six given the man's truthful reporting | Step-by-Step Solution
Problem
A man speaks truth 3 out of 4 times. A die is rolled and the man reports it shows a six. Find the probability that the die actually shows a six.
๐ฏ What You'll Learn
- Apply Bayes' Theorem to solve complex probability problems
- Understand how to calculate conditional probabilities
- Interpret probabilistic reasoning with multiple events
Prerequisites: Basic probability theory, Conditional probability concepts, Probability tree diagrams
๐ก Quick Summary
I can see you're working with a conditional probability problem that involves someone reporting what they observed - this is a perfect setup for Bayes' Theorem! Here's what I'd like you to think about: when this person reports seeing a six, what are ALL the different ways that could have happened? Remember, he could be telling the truth about actually seeing a six, or he could be lying about seeing something else. The key insight is figuring out which scenario is more likely given what we know about his truthfulness and the natural probability of rolling any particular number. Try sketching out these different scenarios and think about how likely each one is - what do you know about the probability of rolling a six, and what do you know about this person's honesty rate?
Step-by-Step Explanation
๐ฒ Let's Solve This Bayes' Theorem Problem Together!
1. What We're Solving:
We need to find the probability that a die actually shows a six, given that a man (who tells the truth 3 out of 4 times) reports seeing a six.2. The Approach:
This is a classic Bayes' Theorem problem! We're dealing with conditional probability - we want to "reverse" our thinking. Instead of asking "If the die shows six, what's the chance he reports it correctly?" we're asking "If he reports six, what's the chance the die actually shows six?"The man could report "six" in TWO scenarios:
- The die IS a six AND he tells the truth
- The die is NOT a six AND he lies
3. Step-by-Step Solution:
Step 1: Define our events
- Let A = "Die actually shows a six"
- Let B = "Man reports a six"
Step 2: Gather our known information
- P(man tells truth) = 3/4, so P(man lies) = 1/4
- P(die shows six) = 1/6, so P(die doesn't show six) = 5/6
- Case 1: Die shows six AND he tells truth = (1/6) ร (3/4) = 3/24
- Case 2: Die shows something else AND he lies = (5/6) ร (1/4) = 5/24
Step 4: Apply Bayes' Theorem P(A|B) = P(B|A) ร P(A) / P(B)
Where:
- P(B|A) = probability he reports six when die shows six = 3/4 (he tells truth)
- P(A) = 1/6 (die shows six)
- P(B) = 1/3 (calculated above)
4. The Answer:
The probability that the die actually shows a six is 3/8 or 0.375 (37.5%)Notice something interesting: Even though he's mostly truthful, the probability is less than 50%! This happens because it's much more likely for the die to show something other than six (5 out of 6 times).
5. Memory Tip:
๐ง Remember: In Bayes' problems, always consider ALL the ways the observed event can happen! The man reporting "six" doesn't just depend on his honesty - it also depends on what the die actually showed. Sometimes the "rare event" (rolling a six) makes the final probability lower than you'd expect!Great job working through this challenging concept! Bayes' Theorem takes practice, but you're building really important statistical reasoning skills! ๐
โ ๏ธ Common Mistakes to Avoid
- Confusing the probability of truth-telling with the final probability
- Incorrectly applying Bayes' Theorem formula
- Misinterpreting the given probability information
This explanation was generated by AI. While we work hard to be accurate, mistakes can happen! Always double-check important answers with your teacher or textbook.

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๐ท Problem detected:
Solve: 2x + 5 = 13
Step 1:
Subtract 5 from both sides...
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