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Guest
The results of a recent census on the number of children in families in Canada are no children: 40%, 1 child 25%, 2 children 18% and 3 or more 17%.
If a Canadian family is selected at random what is the estimated probability that there are more than two children, given that the family has at least one child?
I got the bottom but how did you get the top? The equation used is P(A and B) / P(B), the bottom is at least one child which is obviously P(B) = 60% but the top has to be P(A and B) but how did that turn out to be 17% when P(A) is already 17%
The answer's 17 / 60 = 28.3%
If a Canadian family is selected at random what is the estimated probability that there are more than two children, given that the family has at least one child?
I got the bottom but how did you get the top? The equation used is P(A and B) / P(B), the bottom is at least one child which is obviously P(B) = 60% but the top has to be P(A and B) but how did that turn out to be 17% when P(A) is already 17%
The answer's 17 / 60 = 28.3%