# \[

{\text{There are two coins, one unbiased with probability }}\dfrac{1}{2}{\text{\;}}{\text{\;of getting heads and the other one}} \\

{\text{is biased with probability\;}}\dfrac{3}{4}{\text{\;of getting heads}}{\text{. A coin is selected at random and tossed}}{\text{. }} \\

{\text{It shows heads up}}{\text{. Then, the probability that the unbiased coin was selected is }} \\

{\text{(A)\; }}\dfrac{2}{3}{\text{ (}}{\text{B)\;}}\dfrac{3}{5} \\

({\text{C)\; }}\dfrac{1}{2}{\text{ (}}{\text{D) }}\dfrac{2}{5} \\

\]

Last updated date: 20th Mar 2023

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\[

{\text{Let E}} \to {\text{Event of head showing up}} \\

{\text{Let }}{{\text{E}}_1} \to {\text{Event of biased coin chosen}} \\

{\text{Let }}{{\text{E}}_2} \to {\text{Event of unbiased coin chosen}} \\

{\text{Now as we know that }}P(x){\text{ is the probability of occurrence of x}}{\text{.}} \\

{\text{As the probability of occurrence of }}{E_1}{\text{ and }}{E_2}{\text{ is same so,}} \\

\Rightarrow {\text{Now, }}P({E_2}) = \dfrac{1}{2}{\text{ and }}P({E_1}) = \dfrac{1}{2} \\

{\text{So, by conditional probability we know that }}P\left( {\dfrac{x}{y}} \right){\text{ is the probability of occurrence of x if }} \\

{\text{it is known that y has already occured}} \\

{\text{So as given in question,}} \\

\Rightarrow P\left( {\dfrac{E}{{{E_2}}}} \right) = \dfrac{1}{2}{\text{ and }}P\left( {\dfrac{E}{{{E_1}}}} \right) = \dfrac{3}{4}{\text{ }} \\

{\text{So, now we have to find the probability that coin selected is unbiased if it is known }} \\

{\text{that the coin we get is head showing up}}{\text{.}} \\

{\text{And this is equal to }}P\left( {\dfrac{{{E_2}}}{E}} \right) \\

{\text{So, now by using Baye's theorem}} \\

\Rightarrow {\text{ }}P\left( {\dfrac{{{E_2}}}{E}} \right) = \dfrac{{P({E_2})*P\left( {\dfrac{E}{{{E_2}}}} \right)}}{{P({E_2})*P\left( {\dfrac{E}{{{E_2}}}} \right) + P({E_1})*P\left( {\dfrac{E}{{{E_1}}}} \right)}}{\text{ }}\left( 1 \right) \\

{\text{Putting all values on equation 1 we get,}} \\

\Rightarrow {\text{ }}P\left( {\dfrac{{{E_2}}}{E}} \right) = \dfrac{{\dfrac{1}{2}*\dfrac{1}{2}}}{{\dfrac{1}{2}*\dfrac{1}{2} + \dfrac{1}{2}*\dfrac{3}{4}}}{\text{ = }}\dfrac{{\dfrac{1}{4}}}{{\dfrac{1}{4} + \dfrac{3}{8}}}{\text{ = }}\dfrac{{\dfrac{1}{4}}}{{\dfrac{5}{8}}}{\text{ }} = {\text{ }}\dfrac{2}{5} \\

{\text{So, probability that coin selected is unbiased if it is known that the coin we get is }} \\

{\text{head showing up = }}P\left( {\dfrac{{{E_2}}}{E}} \right) = \dfrac{2}{5} \\

{\text{Hence correct option is D}} \\

{\text{NOTE: - Whenever you came up with this type of problem then you have to apply Baye's}} \\

{\text{theorem}}{\text{. So, you should always remember Baye's theorem}}{\text{.}} \\

\]

{\text{Let E}} \to {\text{Event of head showing up}} \\

{\text{Let }}{{\text{E}}_1} \to {\text{Event of biased coin chosen}} \\

{\text{Let }}{{\text{E}}_2} \to {\text{Event of unbiased coin chosen}} \\

{\text{Now as we know that }}P(x){\text{ is the probability of occurrence of x}}{\text{.}} \\

{\text{As the probability of occurrence of }}{E_1}{\text{ and }}{E_2}{\text{ is same so,}} \\

\Rightarrow {\text{Now, }}P({E_2}) = \dfrac{1}{2}{\text{ and }}P({E_1}) = \dfrac{1}{2} \\

{\text{So, by conditional probability we know that }}P\left( {\dfrac{x}{y}} \right){\text{ is the probability of occurrence of x if }} \\

{\text{it is known that y has already occured}} \\

{\text{So as given in question,}} \\

\Rightarrow P\left( {\dfrac{E}{{{E_2}}}} \right) = \dfrac{1}{2}{\text{ and }}P\left( {\dfrac{E}{{{E_1}}}} \right) = \dfrac{3}{4}{\text{ }} \\

{\text{So, now we have to find the probability that coin selected is unbiased if it is known }} \\

{\text{that the coin we get is head showing up}}{\text{.}} \\

{\text{And this is equal to }}P\left( {\dfrac{{{E_2}}}{E}} \right) \\

{\text{So, now by using Baye's theorem}} \\

\Rightarrow {\text{ }}P\left( {\dfrac{{{E_2}}}{E}} \right) = \dfrac{{P({E_2})*P\left( {\dfrac{E}{{{E_2}}}} \right)}}{{P({E_2})*P\left( {\dfrac{E}{{{E_2}}}} \right) + P({E_1})*P\left( {\dfrac{E}{{{E_1}}}} \right)}}{\text{ }}\left( 1 \right) \\

{\text{Putting all values on equation 1 we get,}} \\

\Rightarrow {\text{ }}P\left( {\dfrac{{{E_2}}}{E}} \right) = \dfrac{{\dfrac{1}{2}*\dfrac{1}{2}}}{{\dfrac{1}{2}*\dfrac{1}{2} + \dfrac{1}{2}*\dfrac{3}{4}}}{\text{ = }}\dfrac{{\dfrac{1}{4}}}{{\dfrac{1}{4} + \dfrac{3}{8}}}{\text{ = }}\dfrac{{\dfrac{1}{4}}}{{\dfrac{5}{8}}}{\text{ }} = {\text{ }}\dfrac{2}{5} \\

{\text{So, probability that coin selected is unbiased if it is known that the coin we get is }} \\

{\text{head showing up = }}P\left( {\dfrac{{{E_2}}}{E}} \right) = \dfrac{2}{5} \\

{\text{Hence correct option is D}} \\

{\text{NOTE: - Whenever you came up with this type of problem then you have to apply Baye's}} \\

{\text{theorem}}{\text{. So, you should always remember Baye's theorem}}{\text{.}} \\

\]

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