a) If the bit string is of length 15 and you want exactly 3 zeroes then..

C(15,3) = 15!/[(15-3)!3!] = 15!/[12!3!] = 455 bit strings of 15 that have exactly three 0's

b) For more 0's than 1's you have to start at the majority of a set of 15...if 7.5 is half of 15 then anything greater than 7.5 would be the majority.

We stick with whole numbers so 8 would be the minimum number of bit strings that have more 0's than 1's. We don't stop here though. We have to account for 8,9,10,11,12...and 15 zeros!