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

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.

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

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" $=\frac{\frac{10}{10}\times \frac{1}{10}}{\frac{1}{100}+\frac{2}{100}+\frac{3}{100}+\frac{4}{100}+\frac{10}{100}}$

$=\frac{10}{55}$

$\frac{2}{11}.$

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,

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

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