We are given three coins: one has heads in both faces, the second has tails in both faces, and the third has a head in one face and a tail in the other. We choose a coin at random, toss it, and the result is heads. What is the probability that the opposite face is tails?

Winston Todd

Winston Todd

Answered question

2022-10-27

Modeling conditional probability
We are given three coins: one has heads in both faces, the second has tails in both faces, and the third has a head in one face and a tail in the other. We choose a coin at random, toss it, and the result is heads. What is the probability that the opposite face is tails?
Solution:
If p = P ( Two headed coin was chosen | Heads came up ) = 1 3 1 2 = 2 3
New to probability, would like to see how this solution works in more detail.
My questions:
1.The sample space is {HH,HT,TT}. The condition that heads came up refers to the elements {HH,HT} and the fact that two headed coin was chosen refers to {HH}. We want the probability of { H H , H T } { H H } = { H H }. The probability of choosing HH is 1 3 by Uniform Probability Law. Is this correct?
2. When we calculate that P ( Heads came up ) = 1 2 , what's the sample space? Is it {H,T} or {HH,HT,TT}?

Answer & Explanation

rcampas4i

rcampas4i

Beginner2022-10-28Added 22 answers

Step 1
For your question 1 you can apply Bayes' theorem by noting that the event you are after is the same as having chosen the coin with both a H and T face, given that you tossed a H:
P ( coin is HT | toss H ) = P ( toss H | coin is HT ) P ( coin is HT ) P ( toss H )
and plugging in the appropriate values.
Step 2
For your question 2 (i.e. the denominator in the above) you can use the Law of Total Probability (LOTP) to partition up the sample space:
P ( toss H ) = P ( toss H | coin is HH ) P ( coin is HH ) + P ( toss H | coin is HT ) P ( coin is HT ) + P ( toss H | coin is TT ) P ( coin is TT )
and evaluating the probabilities.

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