SOLUTION: Binomial Distribution-
15% of the U.S. population is left-handed. Randomly select 10 people. Let x represent the number of left-handers.
a. Find probability that exactly 8 people
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Question 1037745: Binomial Distribution-
15% of the U.S. population is left-handed. Randomly select 10 people. Let x represent the number of left-handers.
a. Find probability that exactly 8 people are left-handed
b. Find probability that at least 1 person is left-handed
I don't understand the formula and how to set it up. I know you can put it into the calculator list but I get weird numbers after 6. Do I stop at 6 when adding the numbers together?
Answer by solver91311(24713) (Show Source): You can put this solution on YOUR website!
The probability of 
successes in 
trials where 
is the probability of success on any given trial is given by:
\ =\ {{n}\choose{k}}\left(p\right)^k\left(1\,-\,p\right)^{n\,-\,k})
Where 
is the number of combinations of things taken at a time and is calculated by !})
You want the probability of 8 successes out of 10 trials where the probability of success on any given trial is 0.15, so:
\ =\ {{10}\choose{8}}\left(0.15\right)^8\left(0.85\right)^{2})
That is just an arithmetic problem.
The probability of at least 1 is the sum of all numbers of successes except 0, so you can use the above process to calculate the probability of exactly 1, then exactly 2, exactly 3, and so on all the way up to 10 and then add all 10 results, OR you can just calculate the probability of exactly 0 successes and subtract that probability from 1.
The choice should be pretty obvious unless you are a glutton for punishment.
\ =\ 1\ -\ P_{10}(0,0.15)\ =\ 1\ -\ {{10}\choose{0}}\left(0.15\right)^0\left(0.85\right)^{10})
Note that 
for any positive integer and 
for all real numbers 
which will simplify your calculation to:
\ =\ 1\ -\ P_{10}(0,0.15)\ =\ 1\ -\ \left(0.85\right)^{10})
John
My calculator said it, I believe it, that settles it
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