Question

There is a fair coin (one side heads, one side tails) and an unfair coin (both sides tails). You pick one at random, flip it 5 times, and observe that it comes up as tails all five times. What is the chance that you are flipping the unfair coin?

Answer 1

Using Bayes' theorem, the chance that you are flipping the unfair coin after observing 5 tails in a row is approximately 96.9%.

Step-by-step explanation:

To solve this problem, we'll use Bayes' theorem, which is a way to find a conditional probability when we have information about reverse conditional probabilities.

Let's denote:

• 
  
• F = picking the fair coin
• 
  
• U = picking the unfair coin
• 
  
• A = observing 5 tails in a row

The probability that we pick the fair coin or the unfair coin is equal since it is stated that we pick one at random. Therefore, P(F) = 0.5 and P(U) = 0.5.

The probability of flipping 5 tails in a row with a fair coin (F) is (1/2)^5 = 1/32 = 0.03125.
Thus, P(A|F) = 0.03125. With an unfair coin (U), since both sides are tails, the probability of flipping 5 tails in a row is 1. So P(A|U) = 1.

Now, let's apply Bayes' theorem to find P(U|A), the probability that we have the unfair coin given that we observed 5 tails in a row:

P(U|A) = [P(A|U) * P(U)] / [P(A|U) * P(U) + P(A|F) * P(F)]

Plugging in the values we have:

P(U|A) = (1 * 0.5) / (1 * 0.5 + 0.03125 * 0.5) = 0.5 / (0.5 + 0.015625) = 0.5 / 0.515625 ≈ 0.969.

The chance that you are flipping the unfair coin is thus approximately 96.9%.

Answer 2

0.9697

Step-by-step explanation:

given that there is a fair coin and an unfair coin.

(both sides tails)

Let B = all tails 5 times

A1- fair coin

A2 - unfair coin

`P(B/A1) = 0.5^5P(B/A2) = 1`

P(A1) = 0.5 and P(A2) =0.5 (since each coin is equally likely to be selected)

Required probability

= Prob (unfair coin/all tails) =P(A2/B)

=
`(P(A2B))/(P(A1B)+P(A2B)) \\=(0.5(1)/(0.5(0.5^5+1)) \\=0.9697`