Question 993825: A Randstad/Harris interactive survey reported that 25% of employees said their company is loyal to them (USA Today, November 11, 2009). Suppose 10 employees are selected randomly and will be interviewed about company loyalty.
What is the probability that at least 2 of the 10 employees will say their company is loyal to them (to 4 decimals)?
The way I tried it was because it is at least 2 so have to do all numbers to 10. so P(2)+P(3)+P(4)+P(5)+P(6)+P(7)+P(8)+P(9)+P(10)= So for this i looked on table adding 1.1664+.0064+.8464+3.6864+8.5264 and so on.. Obviously I am doing this wrong because the answer I put in shows it is incorrect. Can you tell me the formula and how to plug it in by steps? Thanks
Answer by rothauserc(4718)   (Show Source):
You can put this solution on YOUR website! this problem uses the binomial probability(P) formula
P(k successes in n trials) = (combination of n trials with k successes) * p^k * q^(n-k)
for this problem n = 10, k = (2, 3, 4, 5, 6, 7, 8, 9, 10), p = 0.25, q = 0.75
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then P(at least 2 employees say they are loyal) = P(2) + P(3) + P(4) + P(5) + P(6) + P(7) + P(8) + P(9) + P(10) = 0.7559747695922852 approx 0.7560
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