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A survey of 1020 workers in a certain year found that 66% of the respondents spend a total of $40 or less on lunch each week

A survey of 1020 workers in a certain year found that 66% of the respondents spend a total of $40 or less on lunch each week. If 10 of the workers who participated in the survey were chosen at random, what is the probability that at most 3 of them spend a total of $40 or less on lunch each week? (Round your answer to three decimal places.) 

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Answer #1
here this is binomial with parameter n=10 and p=0.66
P(X<=3)= x=0a     (nCx)px(1−p)(n-x)    = 0.022
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