Assume that on a weekday one telephone number out of every ten being called is busy. If 6 randomly selected numbers are called what is the probability that at least three of them be busy?
To calculate the probability that at least three out of six randomly selected numbers will be busy, we can use the binomial probability formula:
P(X ≥ k) = Σ (nCr) * p^k * (1-p)^(n-k)
where:
P(X ≥ k) represents the probability of getting at least k successes,
nCr represents the number of combinations of getting k successes out of n trials,
p represents the probability of success on a single trial, and
(n-k) represents the number of failures on a single trial.
In this case, the probability of getting a busy number on any given call is 1/10, as one out of every ten numbers is busy. Therefore, p = 1/10.
n represents the total number of trials, which is 6 in this case.
Now, we need to calculate the probability for each possible value of k (from 3 to 6) and sum them up.
P(X ≥ 3) = P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6)
P(X = k) = (6Ck) * (1/10)^k * (9/10)^(6-k)
Let's calculate:
P(X = 3) = (6C3) * (1/10)^3 * (9/10)^(6-3) = 20 * (1/1000) * (729/1000) = 0.1458
P(X = 4) = (6C4) * (1/10)^4 * (9/10)^(6-4) = 15 * (1/10000) * (81/100) = 0.0181
P(X = 5) = (6C5) * (1/10)^5 * (9/10)^(6-5) = 6 * (1/100000) * (9/10) = 0.00054
P(X = 6) = (6C6) * (1/10)^6 * (9/10)^(6-6) = 1 * (1/1000000) * 1 = 0.000001
Now, we can calculate the probability of at least three numbers being busy:
P(X ≥ 3) = 0.1458 + 0.0181 + 0.00054 + 0.000001 ≈ 0.164441
Therefore, the probability that at least three out of six randomly selected numbers will be busy is approximately 0.1644, or 16.44%.
To find the probability that at least three out of six randomly selected numbers will be busy, we need to consider the different possible combinations.
First, let's calculate the probability of exactly three, exactly four, exactly five, or exactly six numbers being busy. We can then add up these probabilities to find the overall probability.
To calculate these probabilities, we need to use the binomial probability formula:
P(X = k) = C(n, k) * p^k * (1 - p)^(n - k),
where:
- P(X = k) is the probability of getting exactly k successes (busy phone numbers in this case),
- n is the total number of trials (6 numbers being called),
- k is the number of successful events (busy phone numbers in this case),
- p is the probability of a single success (probability of a phone number being busy), and
- C(n, k) is the number of combinations of n items taken k at a time (n choose k).
Now let's calculate the probabilities for each case.
For exactly three numbers being busy:
P(X = 3) = C(6, 3) * (1/10)^3 * (9/10)^(6 - 3)
For exactly four numbers being busy:
P(X = 4) = C(6, 4) * (1/10)^4 * (9/10)^(6 - 4)
For exactly five numbers being busy:
P(X = 5) = C(6, 5) * (1/10)^5 * (9/10)^(6 - 5)
For all six numbers being busy:
P(X = 6) = C(6, 6) * (1/10)^6 * (9/10)^(6 - 6)
To find the overall probability, we add up these probabilities:
P(at least three numbers being busy) = P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6)
Calculate these individual probabilities, add them together, and you will find the probability of having at least three numbers busy when six numbers are called.