There are three cards. The first is green on both sides, the second is red on bo

Question

There are three cards. The first is green on both sides, the second is red on both sides and the third is green on one side and red on the other. We choose a card at random and we see one side (also chosen at random). If the side we see is green, what is the probability that the other side is also green? Many people intuitively answer 1/2. Show that the correct answer is 2/3. Suppose that a fair coin is tossed repeatedly until both a head and tail have appeared at least once. (a) Describe the sample space Ohm. (b) What is the probability that three tosses will be required? Show that if P(A) = 0 or P(A) = 1 then A is independent of every other event. Show that if A is independent of itself then P(A) is either 0 or 1. The probability that a child has blue eyes is 1/4. Assume independence between children. Consider a family with 3 children. (a) If it is known that at least one child has blue eyes, what is the probability that at least two children have blue eyes? (b) If it is known that the youngest child has blue eyes, what is the probability that at least two children have blue eyes?

Explanation / Answer

Question 12:

Here we are given that there are 3 cards, out of which the first one is green on both sides, second is red on both sides and the third one is green on one side and red on the other side.

Then P( Green | Card 1 ) = 1 that is the probability of getting a green side from card 1.

Similarly, P( Green | Card 2) = 0 and P( Green | Card 2) = 0.5

Now as a card is chosen randomly, P( Card 1 ) = P( Card 2) = P(Card 3) = 1/3

Therefore, from this we get:

P( Green ) = P( Green | Card 1 ) P( Card 1 ) + P( Green | Card 2 ) P( Card 2 ) + P( Green | Card 3 ) P( Card 3)

P( Green ) = (1/3)(1 + 0 + 0.5 ) = 0.5

Now we have to find the probability of getting a card with both green sides given that we got a green, that is:

P( Card 1 | Green )

This could be computed using Bayes formula as:

P( Card 1 | Green ) = P( Green | Card 1 ) P( Card 1 ) / P( Green )

P( Card 1 | Green ) = (1/3)(1) / 0.5 = 2/3

Therefore the required answer here is 2/3

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