Question 1194621: Thirty-five percent of US adults have little confidence in their cars. You randomly select ten US adults. Find the probability that the number of US adults who have little confidence in their cars is (1) exactly six and then find the probability that it is (2) more than 7.
Answer by math_tutor2020(3817)   (Show Source):
You can put this solution on YOUR website!
Part (1)
Computing the probability of exactly 6 people who have little confidence in their cars
The binomial probability formula is
P(x) = (n C x)*(p)^x*(1-p)^(n-x)
where,- n = sample size = 10
- p = 0.35 from the 35%
- x = some value from the set {0,1,2,...,8,9,10} representing the number of people who have little confidence in their cars
- n C x refers to the nCr combination formula notation
For this part, we'll plug in x = 6 to get the following:
P(x) = (n C x)*(p)^x*(1-p)^(n-x)
P(6) = (10 C 6)*(0.35)^6*(1-0.35)^(10-6)
P(6) = (210)*(0.35)^6*(1-0.35)^(10-6)
P(6) = 0.0689098
The value is approximate.
This is about a 6.89% chance of it happening.
Answer: Approximately 0.0689098
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Part (2)
Computing the probability of having more than 7 people who have little confidence in their cars
We need to compute
P(8), P(9), P(10)
Adding them up will get us P(X > 7)
Note that P(7) itself is not part of the sum.
Plug in x = 8
P(x) = (n C x)*(p)^x*(1-p)^(n-x)
P(8) = (10 C 8)*(0.35)^8*(1-0.35)^(10-8)
P(8) = (45)*(0.35)^8*(1-0.35)^(10-8)
P(8) = 0.0042814
I'll let you do the steps for P(9) and P(10)
You should get the following approximate values
P(9) = 0.0005123
P(10) = 0.0000276
Therefore,
P(X > 7) = P(8) + P(9) + P(10)
P(X > 7) = 0.0042814 + 0.0005123 + 0.0000276
P(X > 7) = 0.0048213
Answer: Approximately 0.0048213
Round these answers however your teacher instructs.
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