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Q.A man is known to speak truth 2 out of 3 times. He throws a die and reports that it is a six. Find the probability that it is actually a six. OR An urn contains 4 white and 2 red balls. Find the probability distribution and its mean of the number of red balls, if 2 balls are drawn at random.

Rajasthan RbseRajasthan Board Senior Secondary Examination 2019Subjective· 6mImportance★★★★★
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Use Bayes' theorem with E1=E_1= six occurs, E2=E_2= six does not occur, A=A= reports six. (OR: build the probability distribution of red balls drawn from the urn using the hypergeometric counts, then find its mean.)

Part 1: Let E1=E_1= six actually occurs, P(E1)=16P(E_1)=\dfrac16; E2=E_2= six does not occur, P(E2)=56P(E_2)=\dfrac56.

A=A= the man reports a six. P(A∣E1)=P(A\mid E_1)= probability he speaks the truth =23=\dfrac23. P(A∣E2)=P(A\mid E_2)= probability he lies (and so wrongly reports six) =13=\dfrac13.

By Bayes' theorem:

P(E1∣A)=P(E1)P(A∣E1)P(E1)P(A∣E1)+P(E2)P(A∣E2)=16⋅2316⋅23+56⋅13=2/182/18+5/18=27P(E_1\mid A) = \dfrac{P(E_1)P(A\mid E_1)}{P(E_1)P(A\mid E_1)+P(E_2)P(A\mid E_2)} = \dfrac{\frac16\cdot\frac23}{\frac16\cdot\frac23+\frac56\cdot\frac13} = \dfrac{2/18}{2/18+5/18} = \dfrac{2}{7}

OR: Urn has 4 white + 2 red =6=6 balls; draw 2 at random. Let X=X= number of red balls drawn. Total ways =(62)=15=\binom62=15.

P(X=0)=(42)(20)15=615=25P(X=0)=\dfrac{\binom42\binom20}{15}=\dfrac{6}{15}=\dfrac25

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