Question

Prabhu is known to speak truth 3 out of 4 times. he throws a die and reports that it is a six. Find the probability that it is actually a six?
1. 5/8
2. 7/8
3. 3/8
4. 1/8
5. 1/4

Answer 1

Correct Answer - Option 3 : 3/8

Concept:

Bayes' Theorem:

Let E1, E2, ….., En be n mutually exclusive and exhaustive events associated with a random experiment and let S be the sample space. Let A be any event which occurs together with any one of E1 or E2 or … or En such that P(A) ≠ 0. Then

\(P\left( {{E_i}\;|\;A} \right) = \frac{{P\left( {{E_i}} \right)\; \times \;P\left( {A\;|\;{E_i}} \right)}}{{\mathop \sum \nolimits_{i\; = 1}^n P\left( {{E_i}} \right) \;\times\; P\left( {A\;|\;{E_i}} \right)}},\;i = 1,\;2,\; \ldots .\;,\;n\)

Calculation:

Let X be the event that 'Prabhu reports that six occurs in throwing of the die and let E₁ be the event that six occurs and E₂ be the event that six does not occur.

⇒ P (E₁) = 1/6 and P (E₂) = 1 - P (E1) = 5/6

P (X | E₁) = Probability 'Prabhu speaks truth' = 3/4

P (X | E₂) = Probability 'Prabhu does not speaks truth' = 1 - P (X | E1) = 1/4

Here, we have to find the value of P (E₁ | X).

As we know that according to bayes' theorem: \(P\left( {{E_i}\;|\;A} \right) = \frac{{P\left( {{E_i}} \right)\; \times \;P\left( {A\;|\;{E_i}} \right)}}{{\mathop \sum \nolimits_{i\; = 1}^n P\left( {{E_i}} \right)\;\times\; P\left( {A\;|\;{E_i}} \right)}},\;i = 1,\;2,\; \ldots .\;,\;n\)

\( \Rightarrow P{\rm{(}}{E_1}\;{\rm{|}}X) = \frac{{P\left( {{E_1}} \right) \cdot P\;\left( {X\;|\;{E_1}} \right)}}{{\left[ {P\left( {{E_1}} \right) \cdot P\;\left( {X\;|\;{E_1}} \right)\; + \;P\left( {{E_2}} \right) \cdot P\;\left( {X\;|\;{E_2}} \right)} \right]}}\;\)

\( \Rightarrow P{\rm{(}}{E_1}\;{\rm{|}}X) = \frac{{\frac{1}{6} \cdot \frac{3}{{4}}}}{{\left[ {\frac{1}{6} \cdot \frac{3}{4} + \;\frac{5}{6} \cdot \frac{1}{{4}}} \right]}} = \frac{{3}}{{8}}\;\)