The probability that a randomly selected individual in a certain community has made an online purchase is 0.45. Suppose that a sample of 10 people from the community is selected, what is the probability that at most 3 of them has made an online purchase? Round your answer to 2 decimal places.
Find P(0), P(1), P(2), and P(3). Add together for the total probability.
n = 10
x = 0,1,2,3
p = 0.45
Use a binomial probability table or a calculator. It will be easier than doing this by hand.
If you must do this by hand, here is a formula:
P(x) = (nCx)(p^x)[q^(n-x)]
I'll let you take it from here.
To find the probability that at most 3 people have made an online purchase, we need to calculate the probability of selecting 0, 1, 2, or 3 people who have made an online purchase.
Let's use the binomial probability formula to do this. The formula is given by:
P(x) = (nCx) * p^x * (1-p)^(n-x)
Where:
P(x) is the probability of getting exactly x successes,
n is the total number of trials (sample size),
x is the number of successful outcomes,
p is the probability of success for each trial.
In this case, n = 10 (sample size) and p = 0.45 (probability of making an online purchase).
For x = 0:
P(0) = (10C0) * 0.45^0 * (1-0.45)^(10-0)
For x = 1:
P(1) = (10C1) * 0.45^1 * (1-0.45)^(10-1)
For x = 2:
P(2) = (10C2) * 0.45^2 * (1-0.45)^(10-2)
For x = 3:
P(3) = (10C3) * 0.45^3 * (1-0.45)^(10-3)
Now, let's calculate these probabilities:
P(0) = (10C0) * 0.45^0 * (1-0.45)^(10-0) = 1 * 1 * 0.55^10 ≈ 0.0063
P(1) = (10C1) * 0.45^1 * (1-0.45)^(10-1) = 10 * 0.45 * 0.55^9 ≈ 0.0428
P(2) = (10C2) * 0.45^2 * (1-0.45)^(10-2) = 45 * 0.45^2 * 0.55^8 ≈ 0.1319
P(3) = (10C3) * 0.45^3 * (1-0.45)^(10-3) = 120 * 0.45^3 * 0.55^7 ≈ 0.2324
Now, to find the probability that at most 3 people have made an online purchase, we add the probabilities of these events:
P(at most 3) = P(0) + P(1) + P(2) + P(3) ≈ 0.0063 + 0.0428 + 0.1319 + 0.2324 ≈ 0.4134
Therefore, the probability that at most 3 people have made an online purchase is approximately 0.41 (rounded to 2 decimal places).
To calculate the probability that at most 3 people have made an online purchase, we need to use the binomial distribution formula.
The binomial distribution formula is given by:
P(X = k) = C(n, k) * p^k * (1 - p)^(n - k)
Where:
P(X = k) is the probability of exactly k successes.
n is the number of trials (sample size).
k is the number of successes.
p is the probability of success in a single trial.
In this case, n = 10 (sample size), p = 0.45 (probability of an individual making an online purchase), and we need to find the probability of at most 3 successes (k ≤ 3).
To find the probability of at most 3 successes, we need to calculate three probabilities: P(X = 0), P(X = 1), P(X = 2), and P(X = 3), and then add them together.
P(X = 0) = C(10, 0) * 0.45^0 * (1 - 0.45)^(10 - 0)
P(X = 1) = C(10, 1) * 0.45^1 * (1 - 0.45)^(10 - 1)
P(X = 2) = C(10, 2) * 0.45^2 * (1 - 0.45)^(10 - 2)
P(X = 3) = C(10, 3) * 0.45^3 * (1 - 0.45)^(10 - 3)
To calculate the probabilities, we need to use the combination formula:
C(n, k) = n! / (k! * (n - k)!)
C(10, 0) = 10! / (0! * (10 - 0)!) = 1
C(10, 1) = 10! / (1! * (10 - 1)!) = 10
C(10, 2) = 10! / (2! * (10 - 2)!) = 45
C(10, 3) = 10! / (3! * (10 - 3)!) = 120
After calculating the probabilities, we can add them together to find the probability of at most 3 successes:
P(at most 3 successes) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)
Substitute the calculated values to find the final probability:
P(at most 3 successes) = 1 * 0.45^0 * (1 - 0.45)^10 + 10 * 0.45^1 * (1 - 0.45)^9 + 45 * 0.45^2 * (1 - 0.45)^8 + 120 * 0.45^3 * (1 - 0.45)^7
Now, calculate the final probability using the above formula and round the answer to 2 decimal places.