Question 596202
Use the binomial probability distribution



P(X = 0) = (9 ncr 0)*(0.13)^(0)*(1-0.13)^(9-0) 
P(X = 0) = (9 ncr 0)*(0.13)^(0)*(0.87)^(9-0) 
P(X = 0) = (1)*(0.13)^(0)*(0.87)^9 
P(X = 0) = (1)*(1)*(0.285544154243029527) 
P(X = 0) = 0.285544154243029527


P(X = 1) = (9 ncr 1)*(0.13)^(1)*(1-0.13)^(9-1) 
P(X = 1) = (9 ncr 1)*(0.13)^(1)*(0.87)^(9-1) 
P(X = 1) = (9)*(0.13)^(1)*(0.87)^8 
P(X = 1) = (9)*(0.13)*(0.3282116715437121) 
P(X = 1) = 0.384007655706143157


P(X = 2) = (9 ncr 2)*(0.13)^(2)*(1-0.13)^(9-2) 
P(X = 2) = (9 ncr 2)*(0.13)^(2)*(0.87)^(9-2) 
P(X = 2) = (36)*(0.13)^(2)*(0.87)^7 
P(X = 2) = (36)*(0.0169)*(0.37725479487783) 
P(X = 2) = 0.229521817203671772


P(X = 3) = (9 ncr 3)*(0.13)^(3)*(1-0.13)^(9-3) 
P(X = 3) = (9 ncr 3)*(0.13)^(3)*(0.87)^(9-3) 
P(X = 3) = (84)*(0.13)^(3)*(0.87)^6 
P(X = 3) = (84)*(0.002197)*(0.433626201009) 
P(X = 3) = 0.080024848143808932




In short, 


P(X = 0)= 0.285544154243029527

P(X = 1)= 0.384007655706143157

P(X = 2)= 0.229521817203671772

P(X = 3)= 0.080024848143808932



So 

P(X <= 3) = P(X = 0)+P(X = 1)+P(X = 2)+P(X = 3) 
P(X <= 3) = 0.285544154243029527 + 0.384007655706143157 +0.229521817203671772 + 0.080024848143808932 
P(X <= 3) = 0.979098475296653388



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Answer: 


P(X <= 3) = 0.979098475296653388


Round this to four places to get P(X <= 3) = 0.9791


So the the probability that a survey polling nine people reveals that three or fewer intend to vote for that candidate is <font color="red">0.9791</font>