Q263 / 270 · NDA 2019 (Phase 2) · Mathematics

There are 3 coins in a box. One is a two-headed coin; a fair coin another is; and third is biased coin that comes up heads $75 \%$ of time. When one of the three coins is selected at random and flipped, it shows heads. What is the probability that it was the two-headed coin?

  • A
    $\frac{2}{9}$
  • B
    $\frac{1}{3}$
  • Correct answer
    C
    $\frac{4}{9}$
  • D
    $\frac{5}{9}$

Solution

$\begin{aligned} & \text { } P(T)=\frac{1}{3} ; P\left(\frac{H}{T}\right)=1 \\ & P(F)=\frac{1}{3}, P\left(\frac{H}{F}\right)=\frac{1}{2} \\ & P(B)=\frac{1}{3}, P\left(\frac{H}{B}\right)=\frac{3}{4} \\ & P\left(\frac{T}{H}\right)=\frac{P(T) \cdot P\left(\frac{H}{T}\right)}{P(T) \cdot P\left(\frac{H}{T}\right)+P(F) \cdot P\left(\frac{H}{F}\right)+P(B) \cdot P\left(\frac{H}{B}\right)} \\ & \Rightarrow P\left(\frac{T}{H}\right)=\frac{\frac{1}{3}}{\frac{1}{3}+\frac{1}{6}+\frac{1}{4}} \\ & =\frac{1}{\frac{9}{4}}=\frac{4}{9}\end{aligned}$

Hence, the correct answer is option: 'C'.

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