Question

Suppose that a box contains one fair coin (Heads and Tails are equally likely) and one coin with Heads on each side. Suppose further that someone else selects one of the two coins at random, tosses it three times, and gets the outcome {Heads, Heads, Heads}. In this problem, you will answer the following question in several parts: What is the probability that the chosen coin was the fair coin

Answer 1

Using conditional probability, it is found that there is a 0.1 = 10% probability that the chosen coin was the fair coin.

Conditional Probability

$P(B|A) = \frac{P(A \cap B)}{P(A)}$

In which

• P(B|A) is the probability of event B happening, given that A happened.
• $P(A \cap B)$ is the probability of both A and B happening.
• P(A) is the probability of A happening.

In this problem:

• Event A: Three heads.
• Event B: Fair coin.

The probability associated with 3 heads are:

• $0.5^3 = 0.125$ out of 0.5(fair coin).
• 1 out of 0.5(biased).

Hence:

$P(A) = 0.125 + 0.5 = 0.625$

The probability of 3 heads and the fair coin is:

$P(A \cap B) = 0.5(0.125) = 0.0625$

Then, the conditional probability is:

$P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0.0625}{0.625} = 0.1$

0.1 = 10% probability that the chosen coin was the fair coin.

A similar problem is given at brainly.com/question/14398287