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# There are two coins, one unbiased with probability $\dfrac{1}{2}$ of getting heads and the other one is biased with probability $\dfrac{3}{4}$ of getting heads. A coin is selected at random and tossed. It shows heads up, then the probability that the unbiased coin was selected is A) $\dfrac{2}{3}$B) $\dfrac{3}{5}$C) $\dfrac{1}{2}$D) $\dfrac{2}{5}$

Last updated date: 24th Mar 2023
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Hint: First use conditional probability to find the occurrence of head bases on biased coin and unbiased coin. Later use Bayes theorem

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

Complete step by step solution:

Let E $\to$ Event of head showing up

Let $E_1 \to$ Event of biased coin chosen

Let $E_2 \to$ Event of unbiased coin chosen

Now as we know that P(x) is the probability of occurrence of x.

As the probability of occurrence of $E_1$ and $E_2$ is same so,

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

So, by conditional probability we know that $P\left( {\dfrac{x}{y}} \right)$ is the probability of occurrence of x if it is known that y has already occured.

So as given in question,

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

So, now we have to find the probability that the coin selected is unbiased if it is known that the coin we get is head showing up.

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

So, now by using Bayes theorem

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

Putting all values on equation 1 we get,

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

So, probability that coin selected is unbiased if it is known that the coin we get is

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

Hence the correct option is D.

NOTE: - Whenever we come up with this type of problem, first find the occurence of one event when another event is known by using conditional probability and Bayes theorem.