You need to select 3 people from a population who have a particular blood type, where 20% of the population have that blood type. If you select 10 people, what is the probability that at least 3 have this blood type?
This is how I solved it:
Since there's a 20% chance of success, there is a 80% chance of failure.
I took the probability of 0 successes (no people have this blood type out of our sample of 10), then took this result and subtracted it from one.
1 - (0.80)^10 = 0.893
My question is, did I do this problem correctly? The percentage seems very high, so I'm having doubts.
Thanks
This is how I solved it:
Since there's a 20% chance of success, there is a 80% chance of failure.
I took the probability of 0 successes (no people have this blood type out of our sample of 10), then took this result and subtracted it from one.
1 - (0.80)^10 = 0.893
My question is, did I do this problem correctly? The percentage seems very high, so I'm having doubts.
Thanks
-
Castaway -
This is a binomial distribution problem. You are on the right track with the "at least", but there is a little more that needs to be done:
The probability that at least 3 have this blood type EQUALS:
1 - [ Pr(0) + Pr(1) + Pr(2)]
You did the Pr(0) part, but need to add the other two components before subtracting from 1.
Pr(1) = 10C1 (.2)(.8)^9 , where 10C1 means a combination 10 choose 1
Pr(2) = 10C2 (.2^2)(.8^8)
After doing the math I get 1 - 0.678 = 0.322
Hope that helps
This is a binomial distribution problem. You are on the right track with the "at least", but there is a little more that needs to be done:
The probability that at least 3 have this blood type EQUALS:
1 - [ Pr(0) + Pr(1) + Pr(2)]
You did the Pr(0) part, but need to add the other two components before subtracting from 1.
Pr(1) = 10C1 (.2)(.8)^9 , where 10C1 means a combination 10 choose 1
Pr(2) = 10C2 (.2^2)(.8^8)
After doing the math I get 1 - 0.678 = 0.322
Hope that helps