There are three coins in a barrel. These coins, when flipped, will come up heads with respective probabilities 0.3, 0.5, 0.7. A coin is randomly selected from among these three and is then flipped ten times. Let N be the number of heads obtained on the ten flips.1. Pr[N = 0]2. Pr[N = n] for n = 1, 2, 3,...10.3. If you win $1 each time a head appears and you lose $1 each time a tail appears, is this a fair game? Explain.

Having said this, we are told in the question that the probability of a head is [0.3, 0.5, 0.7]. This implies that the probability of tail in that order is [0.7, 0.5, 0.3].

That is, p= 1 - q.

If we look closely we can see that the two possibilities are a mirror of one another. Meanwhile, only the second coin has an equal probability. However, the bias in the first coin is taken care of by the third coin. This makes it balance.

Now, going by the last injunction stated in the question, if N represents the number of heads obtained on the ten flips.

The first scenario: Pr[N=0]

The second scenario: Pr[N=n], for n = 1,2,3,...,10.

The third scenario: Win $1 each time a head appears and lose $1 each time a tail appears.

Then,

First scenario => Pr[N=0] = $0, lost all!

Second scenario => Pr[N=n] = $10, won all!

Thus, the game is either you win $1 or lose $1 at every flip of the coin.