Question 1201223
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Answer: <font color=red>0.70</font>



Explanation:


We have a binomial process going on because...<ul><li>The seeds either germinate or they don't. There are two outcomes.</li><li>Each seed's germination is independent of any other (assuming one doesn't cross-contaminate another).</li><li>The probability of germination is the same for any seed.</li></ul>n = 9 seeds total = sample size
p = 0.75 = probability of germination
x = number of seeds that germinate.


The value of x is an integer between 0 and 9 inclusive of each endpoint.
The value x is some value in the set {0,1,2,3,4,5,6,7,8,9}


B(x) = binomial probability that exactly x seeds will germinate
B(x) = (n C x)*(p^x)*(1-p)^(n-x)
B(x) = (9 C x)*(0.75^x)*(1-0.75)^(9-x)
B(x) = (9 C x)*(0.75^x)*(0.25)^(9-x)
The first part 9 C x refers to the nCr combination formula. 
Pascal's Triangle can be used to determine the nCr values.


We're asked "what is the probability that <font color=blue>at most 7</font> of them germinate?"
Translation: "What is the probability that <font color=blue>7 or fewer</font> seeds germinate?"


We'll need to compute the summation 
B(0)+B(1)+B(2)+B(3)+B(4)+B(5)+B(6)+B(7)
This is the sum from B(0) to B(7).


Ok that's quite a lot of things to compute and add up.
It would be tedious unnecessary busy-work to expect a student to do this (especially by hand or through use of a pocket calculator).


The much preferred shortcut is to find B(8)+B(9) first. Then subtract that sum from 1.


This works because:
<font color=blue>B(0)+B(1)+B(2)+B(3)+B(4)+B(5)+B(6)+B(7)</font>+<font color=red>B(8)+B(9)</font> = 1
<font color=blue>B(0)+B(1)+B(2)+B(3)+B(4)+B(5)+B(6)+B(7)</font> = 1-(<font color=red>B(8)+B(9)</font>)


Rephrased another way: The events "<font color=blue>at most 7</font>" and "<font color=red>more than 7</font>" are complementary. 
One or the other must happen. 
This is why the terms marked in blue and the terms in red add to 1.


We could rewrite those previous equations by saying
<font color=blue>P(x ≤ 7)</font> + <font color=red>P(x > 7)</font> = 1
<font color=blue>P(x ≤ 7)</font> = 1 - <font color=red>P(x > 7)</font>


Let's compute B(8)
B(x) = (9 C x)*(0.75^x)*(0.25)^(9-x)
B(8) = (9 C 8)*(0.75^8)*(0.25)^(9-8)
B(8) = 9*(0.75^8)*(0.25)^(1)
B(8) = 0.2252540588379
B(8) = 0.22525


Let's compute B(9)
B(x) = (9 C x)*(0.75^x)*(0.25)^(9-x)
B(9) = (9 C 9)*(0.75^9)*(0.25)^(9-9)
B(9) = 1*(0.75^9)*(0.25)^(0)
B(9) = 0.0750846862793
B(9) = 0.07508 


Sum the results
B(8)+B(9) = 0.22525 + 0.07508 = 0.30033


Subtract from 1
1-0.30033 = 0.69967


Therefore,
B(0)+B(1)+B(2)+B(3)+B(4)+B(5)+B(6)+B(7) = 0.69967 approximately.
That then rounds to <font color=red>the final answer 0.70</font> when rounding to two decimal places.


There's roughly a 70% chance that at most 7 seeds (i.e. 7 or fewer) will germinate.


Here are a few binomial distribution calculators.
<a href = "https://www.gigacalculator.com/calculators/binomial-probability-calculator.php">https://www.gigacalculator.com/calculators/binomial-probability-calculator.php</a>
<a href = "https://www.omnicalculator.com/statistics/binomial-distribution">https://www.omnicalculator.com/statistics/binomial-distribution</a>
You could also use a spreadsheet or a TI83/TI84 calculator.


Here is an article talking about binomial probabilities using a TI84
<a href = "https://www.statology.org/binomial-probabilities-ti-84-calculator/">https://www.statology.org/binomial-probabilities-ti-84-calculator/</a>
For this particular problem, we can input <font color=red>binomcdf(9, 0.75, 7)</font> into the TI84. This will add up the values from B(0) to B(7).
Be sure to use the CDF and not the PDF.
The PDF is one specific value, while the CDF adds up multiple values below a specific x value.


Another question involving the binomial probability distribution
<a href = "https://www.algebra.com/algebra/homework/Probability-and-statistics/Probability-and-statistics.faq.question.1201351.html">https://www.algebra.com/algebra/homework/Probability-and-statistics/Probability-and-statistics.faq.question.1201351.html</a>
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