Probability with coin toss Suppose a controversial bill is up for vote in the Senate. In order for a bill to pass, at least 51 of the 100 Senators must vote in favor of it. Currently, the Senate can be broken into three distinct groups: The 27 Senators in group A will vote for the bill. The 34 Senators in group B will vote against the bill. The remaining 39 Senators, making up group C, are completely undecided. In order to decide their votes, each member of group C will toss a fair coin. The result of each Senator’s coin toss is independent of all other tosses. If the toss lands heads, the Senator will vote for the bill. If the toss lands tails, the Senator will vote against the bill.
(a) What is the probability that the bill passes?
(b) Suppose that group B tries to ﬁlibuster the bill, so that it cannot come to vote unless 60 Senators are in favor of it. To counteract this, group A has replaced all of group C’s fair coins with biased coins that come up heads with probability 0.8. What is the probability that the ﬁlibuster is broken (i.e. at least 60 Senators are in favor of the bill)?
 A: If $p$ is the probability that a senator in the group C is voting for the bill, then the probability that at least $24$ of them votes in this way is
$$P_1=\sum_{k=24}^{39} \binom{39}k p^k(1-p)^{34-k}.$$
You want to calculate this for $p=1/2$. So in this case this simplifies to
$$P_1=\frac{\sum\limits_{k=24}^{39} \binom{39}k}{2^{39}}.$$
If you do not want to sum 16 binomial coeffients, you can use symmetry and notice that
$$2\sum_{k=24}^{39} \binom{39}k = \sum_{k=0}^{15} \binom{39}k + \sum_{k=24}^{39} \binom{39}k  = \sum_{k=0}^{39} \binom{39}k - \sum_{k=16}^{23} \binom{39}k = 2^{39} - 2\left(\binom{39}{16} + \binom{39}{17} + \binom{39}{18} + \binom{39}{19}\right),$$
hence
$$P_1=\frac12-\frac{\binom{39}{16} + \binom{39}{17} + \binom{39}{18} + \binom{39}{19}}{2^{39}}.$$

In the second case you need $33$ votes from group C, which means
$$P_2=\sum_{k=33}^{39} \binom{39}k p^k(1-p)^{39-k}.$$
And this time you have $p=4/5$.

As mentioned in the comments, good keyword to learn more about similar problems is binomial distribution. In particular, you can read how this can be approximated in some cases by normal distribution.
