At a major International Airport, 0.49 of the flights arrive on time. A sample of 12 flights is studied. What is the probability that more than 3 of them arrived on time?

Write only a number as your answer. Round to 2 decimal places (for example 0.24). Do not write as a percentage

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0.45

To find the probability that more than 3 flights arrived on time, we will use the binomial probability formula.

P(X > 3) = 1 - P(X <= 3)

Where:
P(X > 3) is the probability that more than 3 flights arrived on time.
P(X <= 3) is the probability that 3 or fewer flights arrived on time.

The binomial probability formula is:

P(x) = nCx * p^x * (1-p)^(n-x)

Where:
n is the number of trials (sample size), which is 12 in this case.
x is the number of successful trials (flights arrived on time).
p is the probability of success (probability that a flight arrives on time), which is 0.49.
(1-p) is the probability of failure (probability that a flight does not arrive on time), which is 1 - 0.49 = 0.51.

Let's calculate P(X <= 3) first:

P(X <= 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)

P(X = x) = nCx * p^x * (1-p)^(n-x)

P(X = 0) = 12C0 * 0.49^0 * 0.51^12 = 1 * 1 * (0.51^12) = 0.006 (rounded to 3 decimal places)
P(X = 1) = 12C1 * 0.49^1 * 0.51^11 = 12 * 0.49 * (0.51^11) = 0.043 (rounded to 3 decimal places)
P(X = 2) = 12C2 * 0.49^2 * 0.51^10 = 66 * (0.49^2) * (0.51^10) = 0.133 (rounded to 3 decimal places)
P(X = 3) = 12C3 * 0.49^3 * 0.51^9 = 220 * (0.49^3) * (0.51^9) = 0.221 (rounded to 3 decimal places)

So, P(X <= 3) = 0.006 + 0.043 + 0.133 + 0.221 = 0.403 (rounded to 3 decimal places).

Now, let's calculate P(X > 3) using the formula mentioned earlier:

P(X > 3) = 1 - P(X <= 3) = 1 - 0.403 = 0.597 (rounded to 3 decimal places).

Therefore, the probability that more than 3 flights arrived on time is 0.597.

To find the probability that more than 3 flights arrived on time, we need to use the binomial probability formula. The formula is:

P(X > k) = 1 - P(X ≤ k)

Where P(X > k) represents the probability of getting more than k successes, P(X ≤ k) represents the probability of getting up to k successes, and X follows a binomial distribution.

In this case, the probability of a flight arriving on time is 0.49, and we have a sample size of 12 flights. We want to calculate the probability of more than 3 flights arriving on time, so k = 3.

First, we need to calculate P(X ≤ 3):

P(X ≤ 3) = sum of the probabilities of getting 0, 1, 2, and 3 successes

We can use the binomial probability formula to calculate each individual probability and sum them up.

P(X = 0) = C(12, 0) * (0.49)^0 * (1 - 0.49)^(12 - 0)
P(X = 1) = C(12, 1) * (0.49)^1 * (1 - 0.49)^(12 - 1)
P(X = 2) = C(12, 2) * (0.49)^2 * (1 - 0.49)^(12 - 2)
P(X = 3) = C(12, 3) * (0.49)^3 * (1 - 0.49)^(12 - 3)

Here, C(n, r) represents the binomial coefficient, which calculates the number of ways to choose r successes from n trials. It is given by:
C(n, r) = n! / (r! * (n - r)!)

Now we can calculate each probability:

P(X = 0) = C(12, 0) * (0.49)^0 * (1 - 0.49)^(12 - 0)
= 1 * 1 * (0.51)^12

P(X = 1) = C(12, 1) * (0.49)^1 * (1 - 0.49)^(12 - 1)
= 12 * (0.49) * (0.51)^11

P(X = 2) = C(12, 2) * (0.49)^2 * (1 - 0.49)^(12 - 2)
= 66 * (0.49)^2 * (0.51)^10

P(X = 3) = C(12, 3) * (0.49)^3 * (1 - 0.49)^(12 - 3)
= 220 * (0.49)^3 * (0.51)^9

Now sum up these probabilities:

P(X ≤ 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)

Finally, calculate the probability of more than 3 flights arriving on time:

P(X > 3) = 1 - P(X ≤ 3)

To find the value of P(X > 3), substitute the calculated values into the formula and round the result to 2 decimal places.