Question

Sixty percent of the eligible voting residents of a certain community support the incumbent candidate. If eight of the residents are selected at random, find the probability that at least five of the eight support the candidate.

Answer 1

By the binomial theorem we know that

$1 = (.4 + .6)^8 \\ = {8 \choose 0} (.4)^{8} (.6)^{0} + {8 \choose 1} (.4)^{7} (.6)^{1} +{8 \choose 2} (.4)^{6} (.6)^{2} + {8 \choose 3} (.4)^{5} (.6)^{3} + {8 \choose 4} (.4)^{4} (.6)^{4} \\ + \quad {8 \choose 5} (.4)^{3} (.6)^{5} + {8 \choose 6} (.4)^{2} (.6)^{6} + {8 \choose 7} (.4)^{1} (.6)^{7} + {8 \choose 8} (.4)^{0} (.6)^{8}$

The probability that exactly 5 of 8 support the incumbent is the term

 ${8 \choose 5} (.4)^{3} (.6)^{5}$

So at least five of eight support is the sum of this term and beyond,

$p={8 \choose 5} (.4)^{3} (.6)^{5} + {8 \choose 6} (.4)^{2} (.6)^{6} + {8 \choose 7} (.4)^{1} (.6)^{7} + {8 \choose 8} (.4)^{0} (.6)^{8}$

No particularly easy way of calculating that except popping it into Wolfram Alpha which reports

$p = \dfrac{ 46413}{78125}$

Shouldn't half the terms work out to .6 ?  Interestingly it's not exactly .6 but pretty close at .594.