Question 602100
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 <font color="red">0.75503872</font>