Suppose that on the average 1 person in every 1000 is an addict. Find the probability that a random sample of 8000 people will yield fewer than 7 addicts.
To find the probability that a random sample of 8000 people will yield fewer than 7 addicts, we can use the binomial distribution.
The binomial distribution is used when we have a fixed number of independent trials, where each trial has the same probability of success. In this case, each person has a probability of being an addict of 1/1000.
The formula to calculate the probability of exactly k successes in n trials is:
P(X = k) = C(n,k) * p^k * (1-p)^(n-k)
Where:
- P(X = k) is the probability of exactly k successes,
- C(n, k) is the number of combinations (n choose k),
- p is the probability of success in a single trial,
- (1-p) is the probability of failure in a single trial,
- n is the number of trials.
Now, we want to find the probability of fewer than 7 addicts in a sample of 8000 people. This means we want to calculate the sum of probabilities for k = 0, 1, 2, 3, 4, 5, and 6.
P(X < 7) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6)
Let's calculate each term of the sum:
P(X = 0) = C(8000, 0) * (1/1000)^0 * (1 - 1/1000)^(8000 - 0)
P(X = 1) = C(8000, 1) * (1/1000)^1 * (1 - 1/1000)^(8000 - 1)
...
P(X = 6) = C(8000, 6) * (1/1000)^6 * (1 - 1/1000)^(8000 - 6)
To calculate each term, we need to evaluate the combination formula C(n, k):
C(n, k) = n! / (k! * (n - k)!)
With these calculations, we can find the probability that a random sample of 8000 people will yield fewer than 7 addicts.