A company's records indicate that 90% are error free,if 8-records are randomly selected, what is the probability that at least 2 records have no errors.
this is a binary probability ... no error / error ... 90% / 10%
"at least two" means not one or zero
(n + e)^8 = n^8 + 8 n^7 e + ... + 8 n e^7 + e^8
subtract the last two terms from 1
... 1 - (8 * .9 * .1^7) - .1^8
To find the probability that at least 2 records out of 8 have no errors, we need to calculate the probability of different scenarios and then sum them up.
The probability that a single record has no error is 90%. Therefore, the probability that a single record has an error is 1 - 0.9 = 0.1.
Now, let's consider the different scenarios:
1. All 8 records have no errors:
P(all 8 have no errors) = (0.9)^8
2. Exactly 1 record has no error:
P(exactly 1 has no errors) = (8 choose 1) * (0.9)^1 * (0.1)^7
3. Exactly 2 records have no errors:
P(exactly 2 have no errors) = (8 choose 2) * (0.9)^2 * (0.1)^6
Since we want to calculate the probability of at least 2 records having no errors, we need to consider the complement of scenarios 1, 2, and 3. The complement of "at least 2" is "none" or "exactly 1".
So, the probability of at least 2 records having no errors is:
P(at least 2 records have no errors) = 1 - P(0 records have no errors) - P(exactly 1 has no errors)
Using the formulas above, we can now calculate the probabilities for each scenario:
P(at least 2 records have no errors) = 1 - (0.1)^8 - (8 choose 1) * (0.9)^1 * (0.1)^7
Calculating this expression will give you the probability that at least 2 out of the 8 randomly selected records have no errors.