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Unit 01 /
Problem 2:


I law of total probability and bayes rule
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Moderate


A box has three coins. One has two heads, one has two tails, and the other is a fair coin with one head and one tail. A coin is chosen at random, is flipped, and comes up heads.
(a) What is the probability that the coin chosen is the two-headed coin?
(b) What is the probability that if the same coin is thrown another time, it will come up heads?

Solution:

(a) What is the probability that the coin chosen is the two-headed coin?

To solve this problem, we can use Bayes' Theorem and the concept of conditional probability.

Let's define the events:

  • $C_{hh}$: The coin chosen is the two-headed coin.
  • $C_{tt}$: The coin chosen is the two-tailed coin.
  • $C_{f}$: The coin chosen is the fair coin.
  • $H$: The coin flip results in heads.

First, we find the probability of getting heads $(P(H))$. Since one coin always lands heads, one always lands tails, and one has a 50% chance of landing heads, $P(H)$ is calculated as follows:

  • The two-headed coin always lands heads: $P(H|C_{hh}) = 1$
  • The two-tailed coin never lands heads: $P(H|C_{tt}) = 0$
  • The fair coin lands heads half the time: $P(H|C_{f}) = 0.5$

Since each coin is equally likely to be chosen, each has a $1/3$ probability of being chosen. Thus, $P(H)$ is:

\begin{align}P(H) &= P(C_{hh}) \cdot P(H|C_{hh}) + P(C_{tt}) \cdot P(H|C_{tt}) + P(C_{f}) \cdot P(H|C_{f}) \\&= \frac{1}{3} \cdot 1 + \frac{1}{3} \cdot 0 + \frac{1}{3} \cdot 0.5 \\&= \frac{1}{3} + 0 + \frac{1}{6} \\&= \frac{1}{2}\end{align}

Now, using Bayes' Theorem, $P(C_{hh}|H)$ is:\begin{align}P(C_{hh}|H) &= \frac{P(H|C_{hh}) \cdot P(C_{hh})}{P(H)} \\& = \frac{1 \cdot \frac{1}{3}}{\frac{1}{2}} \\& = \frac{1}{3} \times 2 \\& = \frac{2}{3}\end{align}

(b) What is the probability that if the same coin is thrown another time, it will come up heads?

Let's define the events:

  • $H_{next}$: heads on next flip.

Given that the coin landed heads the first time, there are two possibilities:

  1. It's the two-headed coin, which will always land heads.
  2. It's the fair coin, which has a 50% chance of landing heads.

We've already calculated that the probability of it being the two-headed coin is 2/3, so the probability of it being fair is the remaining 1/3 (since the two-tailed coin is now impossible). Therefore, the probability of the coin landing heads on the next flip is:

\begin{align}P(H_{next}) &= P(C_{hh}) \cdot P(H|C_{hh}) + P(C_{f}) \cdot P(H|C_{f}) \\&= \frac{2}{3} \cdot 1 + \frac{1}{3} \cdot \frac{1}{2} \\&= \frac{2}{3} + \frac{1}{6} \\& = \frac{4}{6} + \frac{1}{6} \\& = \frac{5}{6}\end{align}