Question

Suppose we have 10 coins such that if the ith coin is flipped, heads will appear with probability i/10, i = 1, 2, ..., 10. When one of the coins is randomly selected and flipped, it shows heads. What is the conditional probability that it was the fifth coin

Short answer

The conditional probability was that it was the fifth coin is$P\left(F_5\mid E\right)=\frac{P\left(E\mid F_5\right)P\left(F_5\right)}{\sum _{i=1}^{10} P\left(E\mid F_i\right)P\left(F_i\right)}$$=\frac{1}{11}.$

Step-by-step solution

Given Information

We have 10 coins such that if the $i^{th}$coin is flipped, heads will appear with probability $\frac{i}{10}$, $i=1,2,...10$

When one of the coins is randomly selected and flipped, it shows heads.

We have to find the conditional probability that it was the fifth coin.

 Calculation of Probability  

Consider $E$being the event the randomly selected coin comes up heads.

Consider$F_i$being the event that the coin was the $i^{th}$coin.

Therefore, localid="1647061945560" $P\left(E/F_i\right)=\frac{i}{10}$for $i=1,2,\dots 10$

And that $P\left(F_i\right)=\frac{1}{10}.$

Calculation of the Conditional Probability of the Tenth Coin

Calculate the conditional probability of the tenth coin

$P\left(E/F_{10}\right)=\frac{10}{10}=1$

$P\left(F_{10}/E\right)=\frac{P\left(F_{/F_{10}}\right)\times P\left(F_{10}\right)}{\sum _{i=1}^{10} \left(P{F}_{F_i}\right)\times P\left(F_i\right)}$

$=\frac{\frac{10}{10}\times \frac{1}{10}}{\sum _{i=1}^{10} \frac{i}{10}\times \frac{1}{10}}$

We get,

role="math" localid="1647062585493" $=\frac{\frac{10}{10}\times \frac{1}{10}}{\frac{1}{100}+\frac{2}{100}+\frac{3}{100}+\frac{4}{100}+\frac{10}{100}}$

$=\frac{10}{55}$

$\frac{2}{11}.$

Calculation of Conditional Probability of Fifth Coin

Now, find the conditional probability that it was the fifth coin.

Using Bayes' rule we have

$P\left(F_5/E\right)=\frac{P\left(E\mid F_5\right)P\left(F_5\right)}{\sum _{i=1}^{10} P\left(E\mid F_i\right)P\left(F_i\right)}$

$=\frac{\left(\frac{5}{10}\right)\left(\frac{1}{10}\right)}{\sum _{i=1}^{10} \frac{i}{10}\left(\frac{1}{10}\right)}$

$=\frac{\frac{5}{100}}{\frac{55}{100}}$

We get,

$=\frac{1}{11}.$

Final Answer

The conditional probability was that it was the fifth coin is$\frac{1}{11}.$