Part A
p = 0.32 = probability of someone voting by internet
n = 13 = sample size
x = 10 = number of people who voted by internet
Apply the binomial probability formula.
[tex]P(x) = (_n C _x)*(p)^{x}*(1-p)^{n-x}\\\\P(10) = (_{13} C _{10})*(0.32)^{10}*(1-0.32)^{13-10}\\\\P(10) = (286)*(0.32)^{10}*(0.68)^{3}\\\\P(10) \approx 0.0010124942242\\\\P(10) \approx \textbf{0.00101}\\\\[/tex]
Note: the [tex]_n C_x[/tex] refers to the combination formula.
Answer: 0.00101
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Part B
We could apply the binomial theorem for the following values of x: {10,11,12,13}. Then would add up the results.
However, such a task is fairly tedious and it's more efficient to use computer software (or a graphing calculator). Also, there's the element of limited time that you'll need to consider. I showed the steps for part A above because your teacher may have wanted you to list them. Usually it's better to rely on software for that task as well. At the very least, it's useful to have software to check your answer.
Whichever method you use, you should find the following:
- P(10) = 0.00101
- P(11) = 0.00013
- P(12) = 0.00001
- P(13) = 0
The last value isn't exactly 0, but it's so small that it's effectively zero. This is because we only have 5 decimal places to work with.
Add up those four values to get:
[tex]P(x \ge 10) = P(10)+P(11)+P(12)+P(13)\\\\P(x \ge 10) \approx 0.00101+0.00013+0.00001+0\\\\P(x \ge 10) \approx 0.00115\\\\[/tex]
A faster shortcut is to use your calculator's binomCDF function.
This result of 0.00115 is fairly small. The usual distinction we make between whether a probability is small or not is to set up some kind of threshold. This threshold is known as the significance level. By default, it's set to 0.05 unless your teacher specifies otherwise. Another common significance level is 0.01, but it's usually mentioned.
Since 0.00115 is smaller than 0.05, and even 0.01, we consider the probability of getting 10 or more internet voters to be a significant event. It's fairly unusual to get 10 or more people voting by internet in this sample of n = 13 people.
Answer: Choice A) Yes, because the probability of 10 or more is 0.00115, which is low.
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Part C
Like with part B, we could find the binomial probability values for x = {1,2,3,...,11,12,13}. We could add up the values like I did, or use the binomCDF function which is faster.
Another approach is to use the complement of the event "at least 1". If x is some number in the set {0,1,2,..,12,13}, then either [tex]x = 0[/tex] or [tex]x \ge 1[/tex]
From this, we can say
[tex]P(x=0) + P(x \ge 1) = 1\\\\P(x \ge 1) = 1-P(x=0)\\\\P(x \ge 1) \approx 1-0.00665\\\\P(x \ge 1) \approx 0.99335\\\\[/tex]
This then rounds to 0.993
Answer: 0.993
