Question 969287: A binomial experiment has 5 trials with p=0.7. What's the probability of getting at least 2 successes?
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! The easiest way to approach this is by finding what you DON'T want and subtracting that from 1. What is the probability of getting 0 or 1 success? Once you have that, then subtract it from 1, and you will have the probability of getting at least two successes.
Probability of 0 successes is 0.3^5 and that is 0.00243.
Probability of 1 success is 5 (the number of ways 1 success can happen)*(0.7)*(0.3^4) The success, 0.7 probability, can be anywhere, and there are 4 failures. Multiplication is commutative, so we just multiply those two products by 5=0.02835.
The sum of 0.00243 and 0.02835 =0.031 (0.03078 to 5 places). BUT, that is what we DON'T want. Two or more successes is everything else, so we subtract 0.031 from 1 to get 0.969 (or 0.96922 to five places).
Many times in statistics, it is easier to find what you don't want and subtract from 1, if it is binomial.
For 2 successes, there are 10 ways, and it would be (0.7^2)(0.3^3)*10=0.1323
For 3 successes, there are 10 ways (5 choose 2 and 5 choose 3 are the same)
(0.7^3)*(0.3*2)*10=0.3087
For 4 successes, it is 5*(0.7^4)(0.3)=0.3602 (rounding)
For 5 successes, it is (0.7^5)=0.16807
Their sum is 0.969, which is the answer above, but with more work.
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