# Probablity calculation for at-least clause case

I saw the question: What is the probability of 1 out of 5 randomly selected people agree with the statement that men having more right for job compare to women?

P(agree) = 0.36
P(disagree) = 1 - .36
p(at least 1 agree) = 1 - 0.74^4 (because 1 is already agree with sentence)


Is that correct?

• Did you find the solution for your problem? Sep 19, 2014 at 21:07

Assuming your sample is being done with replacement, the only experiment that wouldn't result in success is when all 5 sampled people disagree. Thus

$$P(at\space least\space 1\space agree) = 1 - 0.64^5$$

Which yields $0.8926$ . First notice that it is $0.64$ (instead of 0.74) and the power is $5$ (not $4$).

Just to be sure, this little python snippet confirms this result for you:

import numpy as np
import random

def test(pop, samp_size):
return 1 if (1.0 in random.sample(pop, samp_size)) else 0

if __name__== '__main__':
NUM_OF_EXPERIMENTS = 100000;
samp_size = 5
pop = np.concatenate((np.zeros(64000000), np.ones(36000000)))
succes_count = 0.
for i in range(0, NUM_OF_EXPERIMENTS):
succes_count += test(pop, samp_size)

print succes_count / NUM_OF_EXPERIMENTS