Question

A person has 5 coins in his pocket. Two have both sides being heads, one has both sides being tails, and two are normal. The coins cannot be distinguished unless one looks at them.

a) The person closes his eyes, picks a coin from pocket at random, and tosses it. What is the probability that the down-side of the coin is heads?

b) He opens his eyes and sees that the up-side of the coin is heads. What is the probability that the downside is also heads (namely, this is a two-heads coin).

c) Without looking at the other side of the coin, he tosses it again. What is the probability that the downside is heads?

d) Now he looks at the upside of the coin and it is heads. What is the probability that the downside of the coin is heads?

Answer 1

a) Note that *two*, not one, of the coins are normal; the total number of coins is 5, not 4.
So P(H on down side) should be (1/5)(1+1+0+1/2+1/2) = 3/5.

b) P(H on down side given H on upper side)
= P(H on both sides)/P(H on upper side)
= P(H on both sides)/P(H on down side)
= (2/5)/(3/5)
= 2/3, so you answered this part correctly (good job!).

c) From b), we know that, given that the upper side is a head after the first toss, the coin is a two-headed coin with probability 2/3, a two-tailed coin with probability 0, and a normal coin with probability 1/3.
So the probability that this same coin, when tossed again, shows heads on the down side is (2/3)(1) + (0)(0) + (1/3)(1/2) = 5/6.

d) From b), we know that, given that the upper side is a head after the first toss, the coin is a two-headed coin with probability 2/3, a two-tailed coin with probability 0, and a normal coin with probability 1/3.

In this situation, we know from c) that the probability that this same coin, when tossed again, shows heads on the down side is 5/6. Therefore, the probability that this same coin, when tossed again, shows heads on the upper side is also 5/6.

Also in this situation, we already know that the probability that this same coin, when tossed again, shows heads on both sides is 2/3.

Therefore we conclude that, given that the upper side is a head the second time around, the probability that the down side is a head is (2/3)/(5/6) = 4/5.