Q 3.45 - A First Course in Probability

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 

Step-by-Step Solution

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 1: 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. 

Step 2: 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, $P\left(E/F_i\right)=\frac{i}{10}$ for $i=1,2,\dots 10$ 

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

Step 3: 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,

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

$=\frac{10}{55}$

$\frac{2}{11}.$

Step 4: 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,

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

Step 5: Final Answer

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