91Ó°ÊÓ

We can denote the expected number of flips needed as E(N). We can also split this into two cases: the first flip is heads (with probability p) and the first flip is tails (with probability 1-p). After the first flip, we are essentially waiting for the other side to appear. So, we can write the expected value as: E(N) = p(E(N | First flip is heads)) + (1-p)(E(N | First flip is tails)) Now, let's consider the two conditional expected values: E(N | First flip is heads) = 1 + (1-p)E(N), since the first flip is heads and then we expect to need (1-p) more flips before getting a tail. E(N | First flip is tails) = 1 + pE(N), since the first flip is tails and then we expect to need p more flips before getting a head. Now we can plug these values back into the original equation: E(N) = p(1 + (1-p)E(N)) + (1-p)(1 + pE(N)) Now we can solve for E(N): E(N) = 1 + p(1-p)E(N)+ (1-p)(pE(N)) E(N)(1 - p(1-p) - (1-p)p) = 1 E(N) = \(\frac{1}{1 - p(1-p) - (1-p)p}\) E(N) = \(\frac{1}{2p(1-p)}\)

Now, let's consider the ratio of heads to total flips: Ratio of heads = \(\frac{p \cdot E(N)}{E(N)}\) Expected number of flips that land heads = p * E(N) = \(\frac{p}{2p(1-p)}\) = \(\frac{1}{2(1-p)}\)

Similarly, the expected number of flips that land tails can be calculated as: Expected number of flips that land tails = (1-p) * E(N) = \(\frac{1-p}{2p(1-p)}\) = \(\frac{1}{2p}\)

Now, we need to find the expected number of flips needed to get at least two heads and one tail. Let E(M) be the expected number of flips needed to get one more head after getting one head and one tail. E(M) = p(1) + (1-p)(1 + E(M)) E(M) = \(\frac{1}{p}\) Now we can find the expected number of flips needed for two heads and one tail case: Expected number of flips = E(N) + E(M) = \(\frac{1}{2p(1-p)} + \frac{1}{p}\)