A person speaks truth 3 out of 4 times. A dice was rolled in his prese

1.	A person speaks truth 3 out of 4 times. A dice was rolled in his presence and he told that the number was 2, what is the possibility that the number is actually 2?1. \(1\over 6\)2. \(3\over 8\)3. \(1\over 4\)4. \(1\over 8\)
Answer» Correct Answer - Option 2 : \(3\over 8\) Concept: General Rule: The number of ways for selecting r from a group of n (n > r) = nCr The probability of particular case = \(\rm \text{Number of ways for the case can be executed}\over{\text{Total number of ways for selection}}\) Bayes' theorem of probability: If E₁, E₂,…,E_n be a set of events in sample S. Let A be any event associated with S, then according to Bayes theorem, \(\rm P(E_i\|A) = {P(E_i)P(A\|E_i)\over\sum _{k=1}^nP(E_k)P(A\|E_k)}\) Where P(E_i) is the probability of Ei and P(E_i\|A) is the probability that E_i assuming A has already occurred Calculation: In the given case Case A: The person is reported 2 Case E₁: The number on dice is 2 ∴ P(E₁) = \(1\over6\) Case E2: The number on dice is not 2 ∴ P(E2) = \(5\over6\) Man reported 2 and 2 appears , i.e., man speaking true ⇒ P(A\|E1) = \(3\over4\) Man reported 2 and 2 do not appears, i.e., man speaking false ⇒ P(A\|E2) = \(1\over4\) For 2 actually appeared P(E₁\|A) = \(\rm P(A\|E_1)P(E_1)\over P(A\|E_1)P(E_1) + P(A\|E_2) P(E_2)\) ⇒ P(E1\|A) = \(\rm {3\over4}\times{1\over6}\over{3\over4}\times{1\over6} + {1\over4}\times{5\over6}\) ⇒ P(E1\|A) = \(\boldsymbol{3\over8}\)

A person speaks truth 3 out of 4 times. A dice was rolled in his presence and he told that the number was 2, what is the possibility that the number is actually 2?1. \(1\over 6\)2. \(3\over 8\)3. \(1\over 4\)4. \(1\over 8\)

Answer» Correct Answer - Option 2 : \(3\over 8\)

Concept:

General Rule:

The number of ways for selecting r from a group of n (n > r) = nCr
The probability of particular case = \(\rm \text{Number of ways for the case can be executed}\over{\text{Total number of ways for selection}}\)

Bayes' theorem of probability: If E₁, E₂,…,E_n be a set of events in sample S. Let A be any event associated with S, then according to Bayes theorem,

\(\rm P(E_i|A) = {P(E_i)P(A|E_i)\over\sum _{k=1}^nP(E_k)P(A|E_k)}\)

Where P(E_i) is the probability of Ei and P(E_i|A) is the probability that E_i assuming A has already occurred

Calculation:

In the given case

Case A: The person is reported 2

Case E₁: The number on dice is 2

∴ P(E₁) = \(1\over6\)

Case E2: The number on dice is not 2

∴ P(E2) = \(5\over6\)

Man reported 2 and 2 appears , i.e., man speaking true

⇒ P(A|E1) = \(3\over4\)

Man reported 2 and 2 do not appears, i.e., man speaking false

⇒ P(A|E2) = \(1\over4\)

For 2 actually appeared

P(E₁|A) = \(\rm P(A|E_1)P(E_1)\over P(A|E_1)P(E_1) + P(A|E_2) P(E_2)\)

⇒ P(E1|A) = \(\rm {3\over4}\times{1\over6}\over{3\over4}\times{1\over6} + {1\over4}\times{5\over6}\)

⇒ P(E1|A) = \(\boldsymbol{3\over8}\)