SOLUTION: 76% of dog owners buy holiday presentes for their dogs. Suppose n=4 dog owners are randomly selected. Find the probability that three or more buy their dog holiday presents. Rou

Question 602100: 76% of dog owners buy holiday presentes for their dogs. Suppose n=4 dog owners are randomly selected. Find the probability that three or more buy their dog holiday presents. Round to four decimal places
Answer by jim_thompson5910(35256) (Show Source): You can put this solution on YOUR website!
First calculate the individual probabilities P(X = 3) and P(X = 4)

P(X = x) = (n C x)*(p)^(x)*(1-p)^(n-x)
P(X = 3) = (4 C 3)*(0.76)^(3)*(1-0.76)^(4-3)
P(X = 3) = (4 C 3)*(0.76)^(3)*(0.24)^(4-3)
P(X = 3) = (4)*(0.76)^(3)*(0.24)^1
P(X = 3) = (4)*(0.438976)*(0.24)
P(X = 3) = 0.42141696

P(X = x) = (n C x)*(p)^(x)*(1-p)^(n-x)
P(X = 4) = (4 C 4)*(0.76)^(4)*(1-0.76)^(4-4)
P(X = 4) = (4 C 4)*(0.76)^(4)*(0.24)^(4-4)
P(X = 4) = (1)*(0.76)^(4)*(0.24)^0
P(X = 4) = (1)*(0.33362176)*(1)
P(X = 4) = 0.33362176

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Then add them up to get your answer

P(X >= 3) = P(X = 3) + P(X = 4)

P(X >= 3) = 0.42141696 + 0.33362176

P(X >= 3) = 0.75503872

So the probability that three or more buy their dog holiday presents is 0.75503872

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