Math, asked by hfabcjbh8252, 1 year ago

A man is known to speak the truth 3 out of 4 times. He throws a die and reports that it is a six the probability that it is actually 6 is

Answers

Answered by mohnishkrishna05
0

Answer:

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Step-by-step explanation:

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Answered by mathdude500
1

Answer:

\boxed{\sf \: Probability\:that\:it\:actually\:a\:6\:is\:\dfrac{3}{8} \: } \\

Step-by-step explanation:

Let assume the following events

E₁ : die show a number 6.

E₂ : die do not show a number 6

E : a man reports that 6 occurs on the die.

Now,

\sf \: P(E_1) = \dfrac{1}{6}  \\  \\

\sf \: P(E_2) = \dfrac{5}{6} \\  \\

Now,

\sf \: P(E  \: \mid \: E_1) = \dfrac{3}{4}  \\  \\

and

\sf \: P(E  \: \mid \: E_2) = \dfrac{1}{4}  \\  \\

Now, By definition of Bayes Theorem, we have

\sf P(E_1 \mid \: E) = \dfrac{P(E_1) . P(E  \mid E_1)}{P(E_1) . P(E  \mid E_1) + P(E_2) . P(E  \mid E_2)} \\  \\

So, on substituting the values, we get

\sf P(E_1 \mid \: E) = \dfrac{\dfrac{1}{6}  \times \dfrac{3}{4} }{\dfrac{1}{6}  \times \dfrac{3}{4}  + \dfrac{5}{6}  \times \dfrac{1}{4} } \\  \\

\sf P(E_1 \mid \: E) = \dfrac{3}{3 + 5} \\  \\

\sf\implies \sf P(E_1 \mid \: E) = \dfrac{3}{8} \\  \\

Hence,

\sf\implies \sf Probability\:that\:it\:actually\:a\:6\:is\:\dfrac{3}{8} \\  \\

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