From experience, an airline knows that only 80% of the passengers booked for a certain flight actually show up. If 8 passengers are randomly selected, find the probability that fewer than 7 of them show up.
Big fat tiddes
To find the probability that fewer than 7 passengers show up, we need to find the probability that either 6, 5, 4, 3, 2, 1, or 0 passengers show up.
First, let's calculate the probability of exactly k passengers showing up, given that the overall probability of a passenger showing up is 0.8:
P(X = k) = C(n, k) * p^k * (1-p)^(n-k)
where:
- n is the number of passengers selected (8 in this case)
- k is the number of passengers showing up
- p is the probability of a passenger showing up (0.8)
- C(n, k) is the combination of n choose k
Using this formula, let's calculate the probabilities for each case:
Case 0: P(X = 0) = C(8, 0) * (0.8)^0 * (1 - 0.8)^(8 - 0)
Case 1: P(X = 1) = C(8, 1) * (0.8)^1 * (1 - 0.8)^(8 - 1)
Case 2: P(X = 2) = C(8, 2) * (0.8)^2 * (1 - 0.8)^(8 - 2)
Case 3: P(X = 3) = C(8, 3) * (0.8)^3 * (1 - 0.8)^(8 - 3)
Case 4: P(X = 4) = C(8, 4) * (0.8)^4 * (1 - 0.8)^(8 - 4)
Case 5: P(X = 5) = C(8, 5) * (0.8)^5 * (1 - 0.8)^(8 - 5)
Case 6: P(X = 6) = C(8, 6) * (0.8)^6 * (1 - 0.8)^(8 - 6)
Now, calculate each case individually:
Case 0: P(X = 0) = 1 * (0.8)^0 * (1 - 0.8)^8 = 1 * 1 * (0.2)^8 = 0.000016384
Case 1: P(X = 1) = 8 * (0.8)^1 * (1 - 0.8)^7 = 8 * 0.8 * (0.2)^7 = 0.0067108864
Case 2: P(X = 2) = 28 * (0.8)^2 * (1 - 0.8)^6 = 28 * 0.64 * (0.2)^6 = 0.060466944
Case 3: P(X = 3) = 56 * (0.8)^3 * (1 - 0.8)^5 = 56 * 0.512 * (0.2)^5 = 0.201326592
Case 4: P(X = 4) = 70 * (0.8)^4 * (1 - 0.8)^4 = 70 * 0.4096 * (0.2)^4 = 0.301989888
Case 5: P(X = 5) = 56 * (0.8)^5 * (1 - 0.8)^3 = 56 * 0.32768 * (0.2)^3 = 0.25769803776
Case 6: P(X = 6) = 28 * (0.8)^6 * (1 - 0.8)^2 = 28 * 0.262144 * (0.2)^2 = 0.121634816
Now, to find the probability that fewer than 7 passengers show up, we sum the probabilities of each case:
P(X < 7) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6)
= 0.000016384 + 0.0067108864 + 0.060466944 + 0.201326592 + 0.301989888 + 0.25769803776 + 0.121634816
≈ 0.950844383
To find the probability that fewer than 7 passengers show up, we need to consider two cases: 0-6 passengers showing up and all 8 passengers showing up.
Case 1: 0-6 passengers show up.
First, let's find the probability that exactly 0 passengers show up. The probability that a passenger does not show up is 1 - 0.80 = 0.20. Since each passenger's non-arrival is independent of the others, we can multiply the probabilities together:
P(0 passengers show up) = (0.20)^8
Next, let's find the probability that exactly 1 passenger shows up. We have 8 different possible scenarios in which exactly one passenger shows up. For each of these scenarios, we multiply the probability that the passenger shows up (0.80) by the probability that the rest of the passengers do not show up (0.20):
P(1 passenger shows up) = 8 * (0.80)^1 * (0.20)^7
We repeat this calculation for 2, 3, 4, 5, and 6 passengers showing up, respectively, and sum up the probabilities:
P(fewer than 7 passengers show up) = P(0 passengers show up) + P(1 passenger shows up) + P(2 passengers show up) + P(3 passengers show up) + P(4 passengers show up) + P(5 passengers show up) + P(6 passengers show up)
Case 2: All 8 passengers show up.
The probability that all 8 passengers show up is (0.80)^8.
Finally, to find the desired probability, we need to subtract the probability of all 8 passengers showing up from the probability of fewer than 7 passengers showing up:
P(fewer than 7 passengers show up) = P(0-6 passengers show up) - P(all 8 passengers show up)
Plugging in the values and simplifying the calculations will give us the final probability.
Here's one way to do this problem:
n = 8
p = .80
q = 1 - p = 1 - .80 = .20
You will need to find P(7) and P(8). Add those values together, then subtract from 1. This will be your probability.
You can use a binomial probability table, or calculate by hand using the following formula:
P(x) = (nCx)(p^x)[q^(n-x)]
I hope this will help.