Question 548982: Ten percent of the population is left handed. What is the probability that in a group of 10 people, at least two will be left handed?
Answer by mathie123(224)   (Show Source):
You can put this solution on YOUR website! This is a binomial distribution with the P(win)=P(left-handed)=10% and P(fail)=P(right-handed)=90%.
We have a formula we can directly plug into the binomial distribution, but just so we know where it is coming from I will derive it.
P(at least 2 left handed)=1-P(0 left-handed)-P(1 left-handed)
(This takes a bit of thinking... basically everything must add to 100%, so I picked the shorter route and subtracted everything it CANNOT be... please let me know if you are unsure)
Let's first find P(0 lefthanded)
Well this means out of the 10 people, 0 of them are "winning"(i.e. left-handed). how many ways can I choose these 0 people? 10C0=1
Okay so once I have the left-handed people picked out, the rest are just right-handed (10 people).
So the probability that 10 people are right-handed is 90%*90%*...90% (10 times- using product rule)=0.9^(10)
and the probability that 0 people are left-handed is 0.1^0=1
So P(0 left-handed) using the product rule is
(10C0)*(09^(10))*(0.1^0)
I will let you figure this out on your calculator....
We can do a similar thing for P (1 left-handed) and find that in general if we have a "win/fail" situation, and we want the probability that we have r wins out of a total of n trials, our formula will look like this:
P(r wins)=(nCr)*(P(win))^r*(P(fail))^(n-r)
Try this out for P( 1 lefthanded) then put these two values we found into the
P(at least 2 left handed)=1-P(0 left-handed)-P(1 left-handed)
formula we have
Hopefully this helps:)
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