There are three coins in a box. The first coin is two-headed. The second one is a fair coin. The third one is a biased coin that comes up heads $75\%$ of the time. When one of the three coins was picked at random from the box and tossed, it landed heads.

What is the probability that the selected coin was the two-headed coin?

Hint.

Use Bayes’ theorem (Bayes’ rule).

Solution.

Let $E_i$ be the event of the $i$-th coin being picked for $i = 1, 2, 3$. Let $F$ be the event that a coin lands heads.

The required probability can be calculated using Bayes’ rule as follows:
\[P(E_1 \mid F) = \frac{P(F \mid E_1) \cdot P(E_1)}{P(F)}.\] When the two-headed coin is picked, it always lands heads. Thus, we have the conditional probability $P(F \mid E_1) = 1$. The probability that the two-headed coin is selected out of the box is $P(E_1)=1/3$.

The probability in the denominator is calculated using the total probability theorem as follows:
\begin{align*}
P(F) &= \sum_{i=1}^3 P(F \mid E_i) P(E_i) \\[6pt] &= \frac{1}{3}\left(1+\frac{1}{2} + \frac{3}{4} \right)\\[6pt] &= \frac{3}{4}.
\end{align*}

It follows by the formula above that the required probability is
\[P(E_1 \mid F) = \frac{1\cdot \frac{1}{3}}{\frac{3}{4}} = \frac{4}{9}.\] Click here if solved 11

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Let $C$ be the event that a randomly chosen person has lung cancer. Let $S$ be the event of a person being a smoker.
Suppose that 10% of the population has lung cancer and 20% of the population are smokers. Also, suppose that we know that 70% of all people who have lung cancer are smokers.

Then determine the probability of a person having lung cancer given that the person is a smoker.

Solution.

Hint (Bayes’s rule)

Let $E$ and $F$ be events.

Let $P(E \mid F)$ be the probability that $E$ occurs given $F$ occurs. This is called a conditional probability of $E$ given $F$.

Suppose that we know $P(E), P(F)$ and $P(E \mid F)$. Then $P(F \mid E)$ can be computed by Bayes’ theorem (alternatively Bayes’ rule):
\[ P(E \mid F) = \frac{P(E) \cdot P(F \mid E)}{P(F)}.\]

Solution

Given information can be formulated as
\[P(C) = 0.1, P(S) = 0.2, \text{ and } P(S \mid C) = 0.7.\]

The required probability is $P(C \mid S)$. Using Bayes’ rule, we can compute it as follows.
\begin{align*}
P(C \mid S) &= \frac{P(C) \cdot P(S \mid C)}{P(S)}\\[6pt] &= \frac{(0.1)(0.7)}{0.2}\\[6pt] &= 0.35
\end{align*}

Remark

The data given here is artificial for educational purpose and is not based on a scientific fact.

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