SOLUTION: 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

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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 .
Answer by math_tutor2020(3816)   (Show Source): You can put this solution on YOUR website!

Answer: 0.70


Explanation:

We have a binomial process going on because...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 at most 7 of them germinate?"
Translation: "What is the probability that 7 or fewer 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:
B(0)+B(1)+B(2)+B(3)+B(4)+B(5)+B(6)+B(7)+B(8)+B(9) = 1
B(0)+B(1)+B(2)+B(3)+B(4)+B(5)+B(6)+B(7) = 1-(B(8)+B(9))

Rephrased another way: The events "at most 7" and "more than 7" 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
P(x ≤ 7) + P(x > 7) = 1
P(x ≤ 7) = 1 - P(x > 7)

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 the final answer 0.70 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.
https://www.gigacalculator.com/calculators/binomial-probability-calculator.php
https://www.omnicalculator.com/statistics/binomial-distribution
You could also use a spreadsheet or a TI83/TI84 calculator.

Here is an article talking about binomial probabilities using a TI84
https://www.statology.org/binomial-probabilities-ti-84-calculator/
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).
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
https://www.algebra.com/algebra/homework/Probability-and-statistics/Probability-and-statistics.faq.question.1201351.html
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