SOLUTION: A biologist studying a hybrid tomato found that there is a probability of 0.70 that the seeds will germinate. If the biologist plants 10 seeds compute the probability that: a) a

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Question 1194301: A biologist studying a hybrid tomato found that there is a probability of 0.70 that the seeds will germinate. If the biologist plants 10 seeds compute the probability that:
a) at most 7 seeds will germinate
b) at least 4 seeds will germinate
c) between 3 and 7 (inclusive, including 3 and 7) will germinate
d) between 4 and 9 (exclusive) will germinate
e) less than 5 will germinate
f) more than 3 will germinate

Answer by math_tutor2020(3817)   (Show Source): You can put this solution on YOUR website!

n = 10 = sample size
p = 0.70 = probability of germination

We have a binomial distribution problem for the following reasoningWe'll use the aptly named binomial probability formula
P(x) = (n C x)*(p)^x*(1-p)^(n-x)
the values of n and p were mentioned earlier
The n C x refers to the nCr formula

The x values take on items from the set {0,1,2,3,4,5,6,7,8,9,10}
because we'll define x like so
x = number of seeds that germinate

Let's compute P(x) when x = 0
P(x) = (n C x)*(p)^x*(1-p)^(n-x)
P(0) = (10 C 0)*(0.70)^0*(1-0.70)^(10-0)
P(0) = (1)*(0.70)^0*(1-0.70)^(10-0)
P(0) = 0.0000059049

Do the same for x = 1
Keep n = 10 and p = 0.7 the same
P(x) = (n C x)*(p)^x*(1-p)^(n-x)
P(1) = (10 C 1)*(0.70)^1*(1-0.70)^(10-1)
P(1) = (10)*(0.70)^1*(1-0.70)^(10-1)
P(1) = 0.000137781

This process is repeated for the other x values from x = 2 all the way up to x = 10.
Spreadsheet software makes quick work of the computations, and it gives a natural easy way to display it as a table.
xP(x)
00.0000059049
10.000137781
20.0014467005
30.009001692
40.036756909
50.1029193452
60.200120949
70.266827932
80.2334744405
90.121060821
100.0282475249

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The previous section is a lot of set up if you aren't quite familiar with the binomial distribution.
Fortunately, software can be used to quickly generate such a table.

We'll use that table to answer parts (a) through (f)

For part (a), the phrasing "at most" means "that is the highest we can go". It is the ceiling value.
"At most 7" means "7 is the highest we can go".

We're tasked to find P(x ≤ 7)
This is the same as adding P(0) all the way through to P(7)
That's 8 numbers we have to add up.

But we can take a shortcut. Notice that
P(x ≥ 8) = P(8) + P(9) + P(10)
P(x ≥ 8) = 0.2334744405 + 0.121060821 + 0.0282475249
P(x ≥ 8) = 0.3827827864

Then we can say,
P(x ≤ 7) + P(x ≥ 8) = 1
P(x ≤ 7) = 1 - P(x ≥ 8)
P(x ≤ 7) = 1 - 0.3827827864
P(x ≤ 7) = 0.6172172136


Answer: Approximately 0.6172172136
Round this however you need to

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Part (b)

"At least 4" means "4 or more".
We want to compute P(x ≥ 4)

Like before we can use a shortcut to find P(x ≤ 3)
P(x ≤ 3) = P(0) + P(1) + P(2) + P(3)
P(x ≤ 3) = 0.0000059049 + 0.000137781 + 0.0014467005 + 0.009001692
P(x ≤ 3) = 0.0105920784

Then,
P(x ≤ 3) + P(x ≥ 4) = 1
P(x ≥ 4) = 1 - P(x ≤ 3)
P(x ≥ 4) = 1 - 0.0105920784
P(x ≥ 4) = 0.9894079216
The longer alternative method would be to add up P(4) all the way up to P(10) which means you have to add up 7 values.

Answer: Approximately 0.9894079216

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Part (c)

P(3 ≤ x ≤ 7) = P(3) + P(4) + P(5) + P(6) + P(7)
P(3 ≤ x ≤ 7) = 0.009001692 + 0.036756909 + 0.1029193452 + 0.200120949 + 0.266827932
P(3 ≤ x ≤ 7) = 0.6156268272

Answer: Approximately 0.6156268272

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Part (d)

P(4 < x < 9) = P(5 ≤ x ≤ 8)
P(4 < x < 9) = P(5) + P(6) + P(7) + P(8)
P(4 < x < 9) = 0.1029193452 + 0.200120949 + 0.266827932 + 0.2334744405
P(4 < x < 9) = 0.8033426667

Answer: Approximately 0.8033426667

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I'll let you tackle parts (e) and (f)

Hints:
P(x < 5) = P(0) + P(1) + P(2) + P(3) + P(4)
P(x > 3) = P(4) + P(5) + P(6) + P(7) + P(8) + P(9) + P(10)
Or you can take the shortcut that
P(x ≤ 3) + P(x > 3) = 1
where
P(x ≤ 3) = P(0) + P(1) + P(2) + P(3)


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