<PRE><B>Question 1031504</B>
There are 4 possible outcomes for two kids(Y-has heart disease, Z-does not have heart disease)
YY
YZ
ZY
ZZ
With the corresponding probability,
{{{P(YY)=(6/7)(6/7)=36/49}}}
{{{P(YZ)=(6/7)(1/7)=6/49}}}
{{{P(ZY)=(1/7)(6/7)=6/49}}}
{{{P(ZZ)=(1/7)(1/7)=1/49}}}
So,
a){{{P(YY)=36/49}}}
b){{{P(ZZ)=1/49}}}
c) {{{P=P(YZ)+P(ZY)=12/49}}}
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