document.write( "Question 1201223: The manufacturer of a fertilizer guarantees that , with the aid of the fertilizer , 75 % of planted seeds will germinate . Suppose the manufacturer is correct . If 9 seeds planted with the fertilizer are randomly selected , what is the probability that at most 7 of them germinate ? Carry your intermediate computations to at least four decimal places , and round your answer to two decimal places . \n" ); document.write( "
Algebra.Com's Answer #835511 by math_tutor2020(3816)\"\" \"About 
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\n" ); document.write( "Answer: 0.70\r
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\n" ); document.write( "\n" ); document.write( "Explanation:\r
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\n" ); document.write( "\n" ); document.write( "We have a binomial process going on because...
  • The seeds either germinate or they don't. There are two outcomes.
  • Each seed's germination is independent of any other (assuming one doesn't cross-contaminate another).
  • The probability of germination is the same for any seed.
n = 9 seeds total = sample size
\n" ); document.write( "p = 0.75 = probability of germination
\n" ); document.write( "x = number of seeds that germinate.\r
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\n" ); document.write( "\n" ); document.write( "The value of x is an integer between 0 and 9 inclusive of each endpoint.
\n" ); document.write( "The value x is some value in the set {0,1,2,3,4,5,6,7,8,9}\r
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\n" ); document.write( "\n" ); document.write( "B(x) = binomial probability that exactly x seeds will germinate
\n" ); document.write( "B(x) = (n C x)*(p^x)*(1-p)^(n-x)
\n" ); document.write( "B(x) = (9 C x)*(0.75^x)*(1-0.75)^(9-x)
\n" ); document.write( "B(x) = (9 C x)*(0.75^x)*(0.25)^(9-x)
\n" ); document.write( "The first part 9 C x refers to the nCr combination formula.
\n" ); document.write( "Pascal's Triangle can be used to determine the nCr values.\r
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\n" ); document.write( "\n" ); document.write( "We're asked \"what is the probability that at most 7 of them germinate?\"
\n" ); document.write( "Translation: \"What is the probability that 7 or fewer seeds germinate?\"\r
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\n" ); document.write( "\n" ); document.write( "We'll need to compute the summation
\n" ); document.write( "B(0)+B(1)+B(2)+B(3)+B(4)+B(5)+B(6)+B(7)
\n" ); document.write( "This is the sum from B(0) to B(7).\r
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\n" ); document.write( "\n" ); document.write( "Ok that's quite a lot of things to compute and add up.
\n" ); document.write( "It would be tedious unnecessary busy-work to expect a student to do this (especially by hand or through use of a pocket calculator).\r
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\n" ); document.write( "\n" ); document.write( "The much preferred shortcut is to find B(8)+B(9) first. Then subtract that sum from 1.\r
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\n" ); document.write( "\n" ); document.write( "This works because:
\n" ); document.write( "B(0)+B(1)+B(2)+B(3)+B(4)+B(5)+B(6)+B(7)+B(8)+B(9) = 1
\n" ); document.write( "B(0)+B(1)+B(2)+B(3)+B(4)+B(5)+B(6)+B(7) = 1-(B(8)+B(9))\r
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\n" ); document.write( "\n" ); document.write( "Rephrased another way: The events \"at most 7\" and \"more than 7\" are complementary.
\n" ); document.write( "One or the other must happen.
\n" ); document.write( "This is why the terms marked in blue and the terms in red add to 1.\r
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\n" ); document.write( "\n" ); document.write( "We could rewrite those previous equations by saying
\n" ); document.write( "P(x ≤ 7) + P(x > 7) = 1
\n" ); document.write( "P(x ≤ 7) = 1 - P(x > 7)\r
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\n" ); document.write( "\n" ); document.write( "Let's compute B(8)
\n" ); document.write( "B(x) = (9 C x)*(0.75^x)*(0.25)^(9-x)
\n" ); document.write( "B(8) = (9 C 8)*(0.75^8)*(0.25)^(9-8)
\n" ); document.write( "B(8) = 9*(0.75^8)*(0.25)^(1)
\n" ); document.write( "B(8) = 0.2252540588379
\n" ); document.write( "B(8) = 0.22525\r
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\n" ); document.write( "\n" ); document.write( "Let's compute B(9)
\n" ); document.write( "B(x) = (9 C x)*(0.75^x)*(0.25)^(9-x)
\n" ); document.write( "B(9) = (9 C 9)*(0.75^9)*(0.25)^(9-9)
\n" ); document.write( "B(9) = 1*(0.75^9)*(0.25)^(0)
\n" ); document.write( "B(9) = 0.0750846862793
\n" ); document.write( "B(9) = 0.07508 \r
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\n" ); document.write( "\n" ); document.write( "Sum the results
\n" ); document.write( "B(8)+B(9) = 0.22525 + 0.07508 = 0.30033\r
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\n" ); document.write( "\n" ); document.write( "Subtract from 1
\n" ); document.write( "1-0.30033 = 0.69967\r
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\n" ); document.write( "\n" ); document.write( "Therefore,
\n" ); document.write( "B(0)+B(1)+B(2)+B(3)+B(4)+B(5)+B(6)+B(7) = 0.69967 approximately.
\n" ); document.write( "That then rounds to the final answer 0.70 when rounding to two decimal places.\r
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\n" ); document.write( "\n" ); document.write( "There's roughly a 70% chance that at most 7 seeds (i.e. 7 or fewer) will germinate.\r
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\n" ); document.write( "\n" ); document.write( "Here are a few binomial distribution calculators.
\n" ); document.write( "https://www.gigacalculator.com/calculators/binomial-probability-calculator.php
\n" ); document.write( "https://www.omnicalculator.com/statistics/binomial-distribution
\n" ); document.write( "You could also use a spreadsheet or a TI83/TI84 calculator.\r
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\n" ); document.write( "\n" ); document.write( "Here is an article talking about binomial probabilities using a TI84
\n" ); document.write( "https://www.statology.org/binomial-probabilities-ti-84-calculator/
\n" ); document.write( "For this particular problem, we can input binomcdf(9, 0.75, 7) into the TI84. This will add up the values from B(0) to B(7).
\n" ); document.write( "Be sure to use the CDF and not the PDF.
\n" ); document.write( "The PDF is one specific value, while the CDF adds up multiple values below a specific x value.\r
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\n" ); document.write( "\n" ); document.write( "Another question involving the binomial probability distribution
\n" ); document.write( "https://www.algebra.com/algebra/homework/Probability-and-statistics/Probability-and-statistics.faq.question.1201351.html
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