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A box contains coins of which are fair and rest are biased. The probability of getting a head when a fair coin is tossed is 1 / 2, while it is \frac{2}{3} wheh a biased coin is tossed. A coin is drawn from the box at random and is tossed twice. The first time it shows head and the second time it shows tail. If the probability that the coin drawn is fair is \mathrm{f(N, m)}, then the value of \mathrm{43\: f\: (20,12)} must be

Option: 1

25


Option: 2

27


Option: 3

26


Option: 4

24


Answers (1)

best_answer

Let \mathrm{E_1, E_2 \: and \: A} denote the following events

\mathrm{E_1}: coin selected is fair

\mathrm{E_2:} coin selected is biased

\mathrm{A :}

The first toss results in a head and the second toss results in a tail.

\mathrm{ P\left(E_1\right)=\frac{m}{N}, P\left(E_2\right)=\frac{N-m}{N}, P\left(\frac{A}{E_1}\right)=\frac{1}{2} \times \frac{1}{2}=\frac{1}{4} }
and \mathrm{ P\left(\frac{A}{E_2}\right)=\frac{2}{3} \times \frac{1}{3}=\frac{2}{9} }

By Baye's theorem

\mathrm{ P\left(\frac{E_1}{A}\right) =\frac{P\left(E_1\right) \cdot P\left(\frac{A}{E_1}\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)} }

\mathrm{ =\frac{\frac{m}{N} \times \frac{1}{4}}{\frac{m}{N} \times \frac{1}{4}+\left(\frac{N-m}{N}\right) \times \frac{2}{9}} }

\mathrm{ =\frac{9 m}{8 N+m}=f(N, m) \text { (given) } }

\mathrm{ \therefore f(20,12)=\frac{9 \times 12}{8 \times 20+12}=\frac{108}{172}=\frac{27}{43} }

\mathrm{ \Rightarrow \quad 43\: f\: (20,12)=27 }

Hence option 2 is correct.




 

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avinash.dongre

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