Answer :
The total probability of at least five residents supporting the candidate, denoted as P(X≥5), is calculated by summing the probabilities of exactly five, six, seven, and eight residents supporting the candidate.
To find the probability of at least five out of eight randomly selected residents supporting the incumbent candidate when 60% of the community supports them, calculate and sum the binomial probabilities for exactly five to eight residents supporting the candidate.
The student is asking for the probability of at least five out of eight randomly selected community members supporting the incumbent candidate, given that 60% of the eligible voting residents support the candidate. This is a binomial probability problem because each selection is a Bernoulli trial with only two possible outcomes (support or do not support) and the probability of a resident supporting the candidate is constant (60%).
To calculate this probability, we will sum the probabilities of exactly five, six, seven, and eight residents supporting the candidate:
Calculate the probability of exactly 5 residents supporting the candidate using the binomial probability formula: P(X = 5) = (8 choose 5) * (0.6)^5 * (0.4)^3.
Repeat the process for P(X = 6), P(X = 7), and P(X = 8).
Finally, sum these probabilities to get the total probability of at least five residents supporting the candidate: P(X \\u2265 5) = P(X = 5) + P(X = 6) + P(X = 7) + P(X = 8).
This sum provides the required probability.
By the binomial theorem we know that
[tex]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}[/tex]
The probability that exactly 5 of 8 support the incumbent is the term
[tex]{8 \choose 5} (.4)^{3} (.6)^{5}[/tex]
So at least five of eight support is the sum of this term and beyond,
[tex]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}[/tex]
No particularly easy way of calculating that except popping it into Wolfram Alpha which reports
[tex]p = \dfrac{ 46413}{78125}[/tex]
Shouldn't half the terms work out to .6 ? Interestingly it's not exactly .6 but pretty close at .594.