Question 2.34: Each of the 2n members of a committee flips a fair coin in d......
Each of the 2n members of a committee flips a fair coin in deciding whether or not to attend a meeting of the committee; a committee member attends the meeting if an H appears. What is the probability that a majority will show up for the meeting?
There will be majority if there are at least n+ 1 committee members present, which amounts to having at least n+1 H’s in 2n independent throws of a fair coin. If X is the r.v. denoting the number of H’s in the 2n throws, then the required probability is: P(X\geq n+1)={\sum_{x=n+1}^{2n}P(X=x)}.~ However,
P(X=x)={\binom{2n}{x}}\left(\frac{1}{2}\right)^x\left(\frac{1}{2}\right)^{2n-x}={\frac{1}{2^{2n}}}{\binom{2n}{x}},
since there are {\binom{2n}{x}} ways of having x ~ H’s in 2n throws. Therefore
P(X\geq n+1)={\frac{1}{2^{2n}}}\sum\limits_{x=n+1}^{2n}{\binom{2n}{x}}={\frac{1}{2^{2n}}}\left [\sum\limits_{x=0}^{2n}{\binom{2n}{x}}-\sum\limits_{x=0}^{n}{\binom{2n}{x}}\right ]
={\frac{1}{2^{2n}}}{\biggl[}2^{2n}-\sum\limits_{x=0}^{n}{\binom{2n}{x}}{\biggr]}=1-{\frac{1}{2^{2n}}}\sum\limits_{x=0}^{n}{\binom{2n}{x}}.
For example, for 2n=10,P(X\geq6)=1-0.6230=0.377(from the binomial tables).