Question

There are a total three coins. One of the coins is a two headed coin (meaning having head on both faces), another is a biased coin that comes up heads $75\%$ of the times and third is also a biased coin that comes up heads $40\%$ $ of the times. One of these three coins is chosen randomly and tossed and it shows the head. What is the probability that the chosen coin was the two headed coin?

Answer 1

The given question is related to the concept of conditional probability and bayes theorem. In order to solve this question, we first have to find the probability of each event and then use a conditional probability formula to find the required probability.

Formula used:
Conditional probability: $P\left( {\dfrac{{{E_1}}}{A}} \right) = \dfrac{{P\left( {{E_1}} \right)P\left( {\dfrac{A}{{{E_1}}}} \right)}}{{P\left( {{E_1}} \right)P\left( {\dfrac{A}{{{E_1}}}} \right) + P\left( {{E_2}} \right)P\left( {\dfrac{A}{{{E_2}}}} \right) + P\left( {{E_3}} \right)P\left( {\dfrac{A}{{{E_3}}}} \right)}}$

Complete step-by-step solution:
Let the event of selecting the two-headed coin be ${E_1}$ , the event of selecting the biased coin that comes up heads $75\%$ of the times be ${E_2}$ , the event of selecting the biased coin that comes up tails $40\%$ of the times be ${E_3}$ and A be the event of getting head on the coin.
Then, we can say that
$P\left( {{E_1}} \right) = P\left( {{E_2}} \right) = P\left( {{E_3}} \right) = \dfrac{1}{3}$
Therefore,
Probability of getting a head on the coin, given that the coin is two-headed is $P\left( {\dfrac{A}{{{E_1}}}} \right) = 1$
Probability of getting a head on the coin, given that the coin is a biased coin that comes up heads $75\%$ of the time is $P\left( {\dfrac{A}{{{E_2}}}} \right) = \dfrac{{75}}{{100}} = \dfrac{3}{4}$
Probability of getting a head on the coin, given that the coin is a biased coin that comes up tails $40\%$ of the time is $P\left( {\dfrac{A}{{{E_3}}}} \right) = \dfrac{{60}}{{100}} = \dfrac{3}{5}$
By Baye’s theorem, the required probability is
$\Rightarrow P\left( {\dfrac{{{E_1}}}{A}} \right) = \dfrac{{P\left( {{E_1}} \right)P\left( {\dfrac{A}{{{E_1}}}} \right)}}{{P\left( {{E_1}} \right)P\left( {\dfrac{A}{{{E_1}}}} \right) + P\left( {{E_2}} \right)P\left( {\dfrac{A}{{{E_2}}}} \right) + P\left( {{E_3}} \right)P\left( {\dfrac{A}{{{E_3}}}} \right)}} \\\ \Rightarrow \left( {\dfrac{{\dfrac{1}{3} \times 1}}{{\dfrac{1}{3} \times 1}}} \right) + \left( {\dfrac{1}{3} \times \dfrac{3}{4}} \right) + \left( {\dfrac{1}{3} \times \dfrac{3}{5}} \right) \\\ \Rightarrow \dfrac{{20}}{{47}} \\\$

Hence, the required percentage is $\dfrac{{20}}{{47}}$ .
Note: The above question was an easy one. Here, we used the concept of conditional probability. Students should know the topic of conditional probability, so that they can easily solve these types of questions. The area of mistake in these questions is writing wrong values in the formula.