91Ó°ÊÓ

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 $i=1,2,…10$

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

Calculate the conditional probability of the tenth coin

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

$P\left(F_{10}/E\right)=\frac{P\left(F_{/F_{10}}\right)×P\left(F_{10}\right)}{\underset{i=1}{\overset{10}{∑}} \left(P{F}_{F_i}\right)×P\left(F_i\right)}$

$=\frac{\frac{10}{10}×\frac{1}{10}}{\underset{i=1}{\overset{10}{∑}} \frac{i}{10}×\frac{1}{10}}$

We get,

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

$=\frac{10}{55}$

$\frac{2}{11}.$

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

Using Bayes' rule we have

$P\left(F_5/E\right)=\frac{P\left(Eâˆ\pounds F_5\right)P\left(F_5\right)}{\underset{i=1}{\overset{10}{∑}} P\left(Eâˆ\pounds F_i\right)P\left(F_i\right)}$

$=\frac{\left(\frac{5}{10}\right)\left(\frac{1}{10}\right)}{\underset{i=1}{\overset{10}{∑}} \frac{i}{10}\left(\frac{1}{10}\right)}$

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

We get,

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

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