document.write( "Question 1168965: From experience, an airline knows that only 73 percent of the passengers booked on a flight from New York to Los Angeles actually board their flight. If this percentage is correct, what is the probability that, in a random sample of 6 booked passengers from New York to Los Angeles, exactly 3 show up?
\n" ); document.write( "Round your answer to three decimal places.
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Algebra.Com's Answer #793607 by math_tutor2020(3817)\"\" \"About 
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\n" ); document.write( "p = probability of success
\n" ); document.write( "p = probability someone showing up (ie actually boarding their flight)
\n" ); document.write( "p = 0.73\r
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\n" ); document.write( "\n" ); document.write( "n = sample size
\n" ); document.write( "n = 6\r
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\n" ); document.write( "\n" ); document.write( "x = number of people who show up (out of n = 6 people)
\n" ); document.write( "x = 3\r
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\n" ); document.write( "\n" ); document.write( "We'll use the binomial probability distribution formula below
\n" ); document.write( "B(x) = (n C x)*(p^x)*(1-p)^(n-x)
\n" ); document.write( "B(3) = (6 C 3)*(0.73^3)*(1-0.73)^(6-3)
\n" ); document.write( "B(3) = 20*(0.73^3)*(0.27^3)
\n" ); document.write( "B(3) = 20*(0.73^3)*(0.27^3)
\n" ); document.write( "B(3) = 20*0.389017*0.019683
\n" ); document.write( "B(3) = 20*0.007657021611
\n" ); document.write( "B(3) = 0.15314043222
\n" ); document.write( "B(3) = 0.153
\n" ); document.write( "In a sample of 6 people, there's roughly a 15.3% chance of exactly 3 showing up and actually boarding their flight.\r
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\n" ); document.write( "\n" ); document.write( "Note: The notation 6 C 3 refers to the nCr combination notation. You'll use the formula \"nCr+=+%28n%21%29%2F%28r%21%2A%28n-r%29%21%29\" or you can use Pascals Triangle. \r
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\n" ); document.write( "\n" ); document.write( "Final Answer: 0.153\r
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