To determine the probability that at least 2 bags are lost on a randomly selected flight, we can use the concept of binomial probability, where:

• $n = 500$ (the number of flights),
• $p = \frac{20}{500} = 0.04$ (the probability of a bag being lost per flight),
• $q = 1 - p = 0.96$ (the probability of no bag being lost per flight).

However, the problem is asking for the probability of at least 2 bags being lost. This is easier to calculate by first finding the probabilities of 0 and 1 bag being lost, and subtracting their sum from 1.

Step 1: Find the probability that 0 bags are lost.

Using the binomial probability formula: $P(X = k) = \binom{n}{k} p^k (1 - p)^{n - k}$ Where:

• $X$ is the number of lost bags,
• $k$ is the number of successful outcomes (in this case, the number of lost bags),
• $n$ is the number of trials (in this case, flights),
• $p$ is the probability of a success (losing a bag).

For 0 lost bags: $P(X = 0) = \binom{500}{0} (0.04)^0 (0.96)^{500} = 1 \cdot 1 \cdot (0.96)^{500}$ We'll compute this value numerically.

Step 2: Find the probability that 1 bag is lost.

Similarly, for 1 lost bag: $P(X = 1) = \binom{500}{1} (0.04)^1 (0.96)^{499} = 500 \cdot 0.04 \cdot (0.96)^{499}$ We'll compute this value numerically as well.

Step 3: Calculate the probability of at least 2 bags lost.

The probability that at least 2 bags are lost is: $P(X \geq 2) = 1 - P(X = 0) - P(X = 1)$

Now, let's calculate these probabilities.The probability that at least 2 bags are lost on a randomly selected flight is approximately $0.99999997$, or very close to 1 when rounded to four decimal places.

Would you like more details on this, or have any further questions?

Here are 5 related questions:

1. How would the probability change if the sample size increased to 1000 flights?
2. What is the probability of losing exactly 3 bags on a flight?
3. How do we adjust this calculation for non-binomial distributions?
4. Can this method be used for continuous data? Why or why not?
5. What assumptions are made when applying the binomial distribution here?

Tip: Binomial probabilities are great for events with two possible outcomes. If more outcomes exist, consider a multinomial distribution.