A biased coin that lands heads with probability 0.6 is flipped 2 times. a. What is the probability of getting an even number of heads? b. Given that more heads than tails appear, what is the probability that all of the flips are H? c. same as (a) except now the coin is flipped 10 times instead of 2 d. same as (b) except now the coin is flipped 10 times instead of 2 So for this problem I figure there are still 4 possible outcomes for a and b....being HH HT TH TT. For even number of heads, HH or TT must happen. So is the probability = P(HH) + P(TT) / Possible outcomes? and for B....since more Heads is the condition, then P(HH) = 1. C and d I'm not sure how to start since there are 2^10 possible outcomes right?