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  • Solve this problem A bag contains three coins, one of which has head on both sides, another is a biased coin that shows up heads 90% of the time and the third one is an unbiased coin. A coin is taken out from the bag at random and tossed. If it shows

A bag contains three coins, one of which has head on both sides, another is a biased coin that shows up heads 90% of the time and the third one is an unbiased coin. A coin is taken out from the bag at random and tossed. If it shows up a head, then the probability that it is the unbiased coin, is :

  • Option 1)

    \frac{3}{8}

  • Option 2)

    \frac{5}{12}

  • Option 3)

    \frac{5}{24}

  • Option 4)

    \frac{1}{3}

 

Answers (1)

As we learnt in 

BAYE'S Theorem -

If E1, E2, E3......En be n mutually exclusive and exhaustive events and A is an event which occurs together with either E1, E2, E3......En from a portion of the sample space S and A be any event then

P\left ( \frac{E_{k}}{A} \right )= \frac{P\left ( E_{k} \right )\cdot P\left ( \frac{A}{E_{k}} \right )}{P\left ( E_{1} \right )\cdot P\left ( \frac{A}{E_{1}} \right )+P\left ( E_{2} \right )\cdot P\left ( \frac{A}{E_{2}} \right )+\cdots P\left ( E_{k} \right )\cdot P\left ( \frac{A}{E_{k}} \right )}

where S = A_{1}\cup A_{2}\cup \cdots A_{k}  and S is sample space.

- wherein

 

 P(Heads_{A})=1

P(Heads_{B})=0.9

P(Heads_{C})=0.5

P\left ( \frac{Heads}{C} \right )=\frac{\frac{1}{3}\times 0.5}{\frac{1}{3}\times 1+\frac{1}{3}\times 0.9+\frac{1}{3}\times 0.5}=\frac{5}{24}                    


Option 1)

\frac{3}{8}

This option is incorrect

Option 2)

\frac{5}{12}

This option is incorrect

Option 3)

\frac{5}{24}

This option is correct

Option 4)

\frac{1}{3}

This option is incorrect

Posted by

Sabhrant Ambastha

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