document.write( "Question 473251: in a large survey, it was discovered that 1 out of every 5 adults visits disney world every year. if 30 adults are randomly selected, what is the probability that exactly 7 of them will visit disney world this year? \n" ); document.write( "
Algebra.Com's Answer #324537 by jim_thompson5910(35256)\"\" \"About 
You can put this solution on YOUR website!
This is a binomial distribution problem. Recall that the binomial pdf value is given by the formula\r
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\n" ); document.write( "\n" ); document.write( "P(X = x) = (n C x)*(p^x)*(1-p)^(n-x)\r
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\n" ); document.write( "\n" ); document.write( "In this case, n=30, p=1/5=0.2 and x=7.\r
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\n" ); document.write( "\n" ); document.write( "So plug these values into the formula and evaluate\r
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\n" ); document.write( "\n" ); document.write( "P(X = 7) = (30 C 7)*(0.2)^(7)*(1-0.2)^(30-7) \r
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\n" ); document.write( "\n" ); document.write( "P(X = 7) = (30 C 7)*(0.2)^(7)*(0.8)^(30-7) \r
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\n" ); document.write( "\n" ); document.write( "P(X = 7) = (2035800)*(0.2)^(7)*(0.8)^23 \r
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\n" ); document.write( "\n" ); document.write( "Note: 30 C 7 = (30!)/(7!(30-7)!) = 2035800\r
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\n" ); document.write( "\n" ); document.write( "P(X = 7) = (2035800)*(0.0000128)*(0.00590295810358705651712) \r
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\n" ); document.write( "P(X = 7) = 0.1538206989732163796166770688\r
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\n" ); document.write( "\n" ); document.write( "P(X = 7) = 0.1538\r
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\n" ); document.write( "\n" ); document.write( "Question: what is the probability that exactly 7 of them will visit disney world this year?\r
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\n" ); document.write( "\n" ); document.write( "Answer: Probability is approximately 0.1538, which is roughly 15.38%
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