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Mathematics 22 Online
OpenStudy (anonymous):

in a family of two children, if the probabilities of a male child and a female child are both 1/2, are the events: "each child is the same sex" and "at most one male" independent?

OpenStudy (anonymous):

P(each child is the same sex and at most one male)=0 P(at most one male)=1/2 So they are not independent

OpenStudy (kropot72):

Independent events have no effect on the occurrence (or non-occurrence) of each other. If one child is a male then the event "each child is the same sex" cannot occur. Therefore the events are not independent.

OpenStudy (anonymous):

i am pretty sure this statement P(each child is the same sex and at most one male)=0 is wrong

OpenStudy (anonymous):

in fact i am sure it is if there are two girls, each child is of the same sex and there is at most one male (since there is none)

OpenStudy (anonymous):

this is not a hard sample space to write out, there are only 4 outcomes, each equally likely (b,b), (b, g) (g, b), (g,g) the probability of each is \(\frac{1}{4}\)

OpenStudy (anonymous):

the probability that both children are of the same sex is \(\frac{1}{2}\) and the probability that there is at most one male \(\frac{3}{4}\)

OpenStudy (anonymous):

put A as the event there is at most one male, B as the event the children are of the same sex, then compute \[\frac{P(A\cap B)}{P(B)}\] and see if it is the same as \(P(A)\)

OpenStudy (anonymous):

\[P(A)=\frac{3}{4}\] \[P(B)=\frac{1}{2}\] \[P(A\cap B)=\frac{1}{4}\] the last means they are both girls

OpenStudy (anonymous):

then since \[\frac{P(A\cap B)}{P(B)}=\frac{\frac{1}{4}}{\frac{1}{2}}=\frac{1}{2}\neq P(A)=\frac{3}{4}\] this tells you they are DEPENDENT

OpenStudy (anonymous):

Oops sorry I've made a serious mistake there, just follow what satellite73 has said!

OpenStudy (anonymous):

thanks everyone. do you know if it would be independent for a three child family? would i be able to find the answer by creating a new sample spacce and doing the same method?

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