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 reasoning- There are two outcomes: Either a seed germinates, or it does not.
- Each trial (seed) is independent of one another. No one seed affects the germination of another.
- The probability of success (ie germination) is the same for each trial, which is p = 0.70
- There are a fixed number of trials (n = 10).
We'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.
| x | P(x) | | 0 | 0.0000059049 | | 1 | 0.000137781 | | 2 | 0.0014467005 | | 3 | 0.009001692 | | 4 | 0.036756909 | | 5 | 0.1029193452 | | 6 | 0.200120949 | | 7 | 0.266827932 | | 8 | 0.2334744405 | | 9 | 0.121060821 | | 10 | 0.0282475249 |
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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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