document.write( "Question 1163614: A hospital spokesperson reported that 4 births had taken place at the Taguig-Pateros Hospital
\n" );
document.write( "during the last 24 hrs. Find the following probabilities:
\n" );
document.write( "a. P(A) = P(the event that 2 boys and 2 girls are born)
\n" );
document.write( "b. P(B) = P(no boys are born)
\n" );
document.write( "c. P(C) = P(at least one boy is born)
\n" );
document.write( "d. P(A|C)
\n" );
document.write( "e. P(B|C)
\n" );
document.write( "f. Are A and C mutually exclusive events? Are A and C independent?
\n" );
document.write( "g. Are B and C mutually exclusive events? Are B and C independent? \n" );
document.write( "
Algebra.Com's Answer #787868 by Boreal(15235)![]() ![]() You can put this solution on YOUR website! 1--4--6--4--1 \n" ); document.write( "2 boys and 2 girls is 6/16 or 0.375 probability \n" ); document.write( "no boys is 1/16 or 0.0625 probability \n" ); document.write( "at least one boy is born is the complement or 0.9375 probability \n" ); document.write( "given at least 1 boy, probability of two boys \n" ); document.write( "this would be 6/15, since given at least one boy decreases denominator from 16 to 15. \n" ); document.write( "P(B|C) is 0. \n" ); document.write( "A and C are not mutually exclusive, B and C are mutually exclusive\r \n" ); document.write( "\n" ); document.write( "P(A)*P(C)= P(A and C) \n" ); document.write( "(5/16)(15/16)=75/256; P(A and C) =P(A)=5/16 \n" ); document.write( "They aren't equal so are dependent. \n" ); document.write( "Once you know 2 boys and 2 girls, you know there is at least one boy. \n" ); document.write( " |