document.write( "Question 1031504: Records show that a child of parents with heart disease has a probability of (fraction) 6/7 of inheriting the disease. Assuming independence, what is the probability that, for a couple with heart disease that have two children: \r
\n" ); document.write( "\n" ); document.write( "(a) Both children have heart disease. Enter the exact answer.
\n" ); document.write( "(b) Neither child has heart disease. Enter the exact answer.
\n" ); document.write( "(c) Exactly one child has heart disease. Enter the exact answer
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Algebra.Com's Answer #646289 by Fombitz(32388)\"\" \"About 
You can put this solution on YOUR website!
There are 4 possible outcomes for two kids(Y-has heart disease, Z-does not have heart disease)
\n" ); document.write( "YY
\n" ); document.write( "YZ
\n" ); document.write( "ZY
\n" ); document.write( "ZZ
\n" ); document.write( "With the corresponding probability,
\n" ); document.write( "\"P%28YY%29=%286%2F7%29%286%2F7%29=36%2F49\"
\n" ); document.write( "\"P%28YZ%29=%286%2F7%29%281%2F7%29=6%2F49\"
\n" ); document.write( "\"P%28ZY%29=%281%2F7%29%286%2F7%29=6%2F49\"
\n" ); document.write( "\"P%28ZZ%29=%281%2F7%29%281%2F7%29=1%2F49\"
\n" ); document.write( "So,
\n" ); document.write( "a)\"P%28YY%29=36%2F49\"
\n" ); document.write( "b)\"P%28ZZ%29=1%2F49\"
\n" ); document.write( "c) \"P=P%28YZ%29%2BP%28ZY%29=12%2F49\"
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