The Probability that a randomly selected commuter listens to audiobooks during their commute is 0.49. Suppose that a sample of 10 commuters is selected, what is the probability that at most 3 of them listen to audiobooks?
Write only a number as your answer. Round to 2 decimal places (for example 0.24). Do not write as a percentage.
pls help!
To solve this problem, we need to use the binomial probability formula. The formula for the probability of exactly x successes in n trials is:
P(x) = C(n, x) * p^x * (1-p)^(n-x)
Where:
P(x) - probability of x successes
C(n, x) - combination formula (n choose x), which calculates the number of ways to choose x elements from a set of n elements
p - probability of success in a single trial
n - number of trials
In this case, the probability of a commuter listening to audiobooks is 0.49, so p = 0.49. We want to find the probability that at most 3 out of 10 commuters listen to audiobooks, which means we need to calculate the probabilities for x = 0, 1, 2, and 3, and then sum them up.
Let's calculate each of these probabilities using the formula:
P(0) = C(10, 0) * 0.49^0 * (1-0.49)^(10-0)
P(1) = C(10, 1) * 0.49^1 * (1-0.49)^(10-1)
P(2) = C(10, 2) * 0.49^2 * (1-0.49)^(10-2)
P(3) = C(10, 3) * 0.49^3 * (1-0.49)^(10-3)
Then, we sum up these probabilities:
P(at most 3) = P(0) + P(1) + P(2) + P(3)
Finally, we round the answer to 2 decimal places.
Now, let's calculate the answer step-by-step:
P(0) = C(10, 0) * 0.49^0 * (1-0.49)^(10-0) = 1 * 1 * 0.51^10 = 0.0219
P(1) = C(10, 1) * 0.49^1 * (1-0.49)^(10-1) = 10 * 0.49 * 0.51^9 = 0.1151
P(2) = C(10, 2) * 0.49^2 * (1-0.49)^(10-2) = 45 * 0.49^2 * 0.51^8 = 0.2451
P(3) = C(10, 3) * 0.49^3 * (1-0.49)^(10-3) = 120 * 0.49^3 * 0.51^7 = 0.2945
P(at most 3) = P(0) + P(1) + P(2) + P(3) = 0.0219 + 0.1151 + 0.2451 + 0.2945 = 0.6766
Rounded to 2 decimal places, the probability that at most 3 out of 10 commuters listen to audiobooks is 0.68.