document.write( "Question 1132664: A researcher found that 55% of families in the United States have one child under the age of 21, 22% have two children under the age of 21, 15% have three children under the age of 21, 5% have 4 children under the age of 21and 3% have 5 or more children under the age of 21. If a family is selected at random, what is the probability the family has at least 3 children under the age of 21? (Express your answer as a percent, Please include the % symbol.)
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Algebra.Com's Answer #749758 by Theo(13342)\"\" \"About 
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i would think it would be the probability of having exactly 3 plus exactly 4 plus exactly 5 or more.
\n" ); document.write( "that would be equal to 15 + 5 + 3 = 23%.\r
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\n" ); document.write( "\n" ); document.write( "to test this out, assume 100 families that fit the proportions shown.
\n" ); document.write( "55 would have one child under 21.
\n" ); document.write( "22 would have two children under 21.
\n" ); document.write( "15 would have three children under 21.
\n" ); document.write( "5 would have four children under 21.
\n" ); document.write( "3 would have 5 or more children under the age of 21.\r
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\n" ); document.write( "\n" ); document.write( "count up the families that have 3 or more children under 21 and you get 15 + 5 + 3 = 23.\r
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\n" ); document.write( "\n" ); document.write( "if you pick a family out of the 100 at random, 23 times out of 100 you will get a family that has at least 3 children.\r
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\n" ); document.write( "\n" ); document.write( "that's a probability of 23%.\r
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