A box contains three coins, one of which is fair, one double-headed (i.e., heads on both sides), and the third is biased in such a way that it comes up heads with probability 3/4. A coin is drawn at random from the box and flipped twice. If both flips result in heads, what is the probability that the coin drawn was double-headed?

#### Solution

Let $A_1$ denote the event that the selected coin is double-headed, let $A_2$ denote the event that the coin is fair, let $A_3$ denote the event that the coin is biased, and let B denote the event that two heads is obtained. Then

` $$ \begin{equation*} P(A_1) = \frac{1}{3}, P(A_2) = \frac{1}{3}, P(A_3) =\frac{1}{3}. \end{equation*} $$ `

` $$ \begin{eqnarray*} P(B|A_1) & = & 1\times 1=1 \quad \because \text{the coin is double-headed} \\ P(B|A_2) & = &\frac{1}{2}\times \frac{1}{2}= \frac{1}{4}\quad \because \text{the coin is fair}\\ P(B|A_3) & = &\frac{3}{4}\times\frac{3}{4}=\frac{9}{16}\quad \because \text{ the coin is biased}. \end{eqnarray*} $$ `

We want to find the probability $P(A_1|B)=\frac{P(A_1\cap B)}{P(B)}$.

The numerator is

` $$ \begin{equation*} P(A_1\cap B)=P(A_1)P(B|A_1)=\frac{1}{3}(1)=\frac{1}{3} \end{equation*} $$ `

and the denominator is

` $$ \begin{eqnarray*} P(B) &=& P(A_1\cap B)+P(A_2\cap B)+ P(A_3\cap B) \\ &=& P(A_1)P(B|A_1) + P(A_2)P(B|A_2) + P(A_3)P(B|A_3)\\ &=& \bigg(\frac{1}{3} \times (1)\bigg) + \bigg(\frac{1}{3} \times \frac{1}{4}\bigg) + \bigg(\frac{1}{3} \times\frac{9}{16}\bigg)\\ &=& \frac{1}{3}\times \frac{29}{16} \end{eqnarray*} $$ `

So, if both flips result in heads, the probability that the coin drawn was double-headed

` $$ \begin{equation*} P(A_1|B)=\frac{P(A_1\cap B)}{P(B)}=\frac{\frac{1}{3}}{\frac{1}{3}\times\frac{29}{16}}=\frac{16}{29}. \end{equation*} $$ `

#### Further Reading

- Statistics
- Descriptive Statistics
- Probability Theory
- Probability Distribution
- Hypothesis Testing
- Confidence interval
- Sample size determination
- Non-parametric Tests
- Correlation Regression
- Statistics Calculators