Question

BlueSky Air has the best on-time arrival rate with 80% of its flights arriving on time. A test is conducted by randomly selecting 16 BlueSky Air flights and observing whether they arrive on time.

What is the probability that 2 flights of BlueSky Air arrive late? (Please carry answers to at least six decimal places in intermediate steps. Give your final answer to the nearest four decimal places).

Probability (as a proportion) ??

Answer 1

0.211 = 21.1% probability that 2 flights of BlueSky Air arrive late

Explanation:

For each flight, there are only two possible outcomes. Either they arrive late, or they do not. The probability of a flight arriving late is independent of any other flight. This means that the binomial probability distribution is used to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

In which is the number of different combinations of x objects from a set of n elements, given by the following formula.

And p is the probability of X happening.

BlueSky Air has the best on-time arrival rate with 80% of its flights arriving on time.

This means that 100 - 80 = 20% are late, which means that

Randomly selecting 16 BlueSky Air flights

This means that

What is the probability that 2 flights of BlueSky Air arrive late?

This is P(X = 2).

0.211 = 21.1% probability that 2 flights of BlueSky Air arrive late

Answer 2

P(X = 2) = 0.2111 (4 d.p.)

Explanation:

To calculate the probability that exactly two out of 16 BlueSky Air flights arrive late, we can use the binomial probability formula:

`\boxed{\displaystyle P(X = k) = \binom{n}{k} \cdot p^k \cdot (1 - p)^(n - k)}`

where:

• n is the total number of trials (16 flights in this case).
• k is the number of successful trials (2 flights arriving late).
• p is the probability of success on each trial (probability of a flight arriving late).
• (1 - p) is the probability of failure on each trial (probability of a flight arriving on time).

Since we want to find the probability of flights arriving late, we consider "arriving late" as a success and "arriving on time" as a failure, so we have:

• n = 16
• k = 2
• p = 1 - 0.80 = 0.20

Substitute these values into the formula:

`\displaystyle P(X = 2) = \binom{16}{2} \cdot (0.20)^2 \cdot (1 - 0.20)^(16 - 2)`

`\displaystyle P(X = 2) = (16!)/(2!(16-2)!) \cdot (0.20)^2 \cdot (0.80)^(14)`

`\displaystyle P(X = 2) = (16!)/(2!\:14!) \cdot 0.04 \cdot 0.043980...`

`\displaystyle P(X = 2) = 120 \cdot 0.04 \cdot 0.043980...`

`\displaystyle P(X = 2) = 4.8 \cdot 0.043980...`

`P(X=2)=0.21110623...`

`P(X=2)=0.2111\; \sf (4\;d.p.)`

Therefore, the probability that exactly 2 flights out of 16 BlueSky Air flights arrive late is 0.2111, rounded to four decimal places.