Isaac, Jennifer, and Karl are making their airline reservations to fly to Kennedy airport (NYC) from Indianapolis on the day after the final exam. All three want to fly on UV Air. UV Air has morning flights from Indianapolis to Kennedy at 7:30 am, 8:15 am, 9:30 am, and 10:20 am, and afternoon flights at 12:20 pm, 1:20 pm, 2:00 pm, 3:30 pm, and 5:00 pm. There are still lots of seats left on each flight, so each may reserve a flight without concern for the others.
How many overall outcomes are possible?
In how many of these outcomes are all three of them on the same flight?
In how many of these outcomes is exactly one of them on the 8:15 am flight?
In how many of these outcomes would exactly one of them have a morning flight?
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To find the answers to these questions, we need to use the concept of combinations.
1. How many overall outcomes are possible?
To calculate the total number of overall outcomes, we need to find the number of options for each person and multiply them together.
Each person can choose from any of the flights, so for Isaac, there are 9 choices, for Jennifer also 9 choices, and for Karl again 9 choices.
To find the total number of outcomes, we multiply the number of choices for each person: 9 * 9 * 9 = 729 outcomes.
2. How many outcomes have all three of them on the same flight?
Since all three individuals need to be on the same flight, we have a limited number of options. There are 9 flights to choose from, so for this question, there is only one possible outcome.
3. How many outcomes have exactly one of them on the 8:15 am flight?
In this case, we will consider three separate scenarios since each person can be the one on the 8:15 am flight.
a) If Isaac is on the 8:15 am flight, Jennifer and Karl have 9 choices each for their flights (excluding 8:15 am, as Isaac has taken that flight). So, for this scenario, there are 9 * 8 * 8 = 576 outcomes.
b) If Jennifer is on the 8:15 am flight, Isaac and Karl have 9 choices each for their flights (excluding 8:15 am). So, for this scenario, there are also 9 * 8 * 8 = 576 outcomes.
c) If Karl is on the 8:15 am flight, Isaac and Jennifer have 9 choices each for their flights (excluding 8:15 am). So, for this scenario, there are 9 * 8 * 8 = 576 outcomes.
To find the total number of outcomes where exactly one of them is on the 8:15 am flight, we add the number of outcomes for each scenario: 576 + 576 + 576 = 1728 outcomes.
4. How many outcomes have exactly one of them on a morning flight?
For this question, we need to consider once again three separate scenarios because each person can be the one on a morning flight.
a) If Isaac is on a morning flight, Jennifer and Karl have 4 choices each for their flights (from 7:30 am, 9:30 am, and 10:20 am). So, for this scenario, there are 4 * 4 * 4 = 64 outcomes.
b) If Jennifer is on a morning flight, Isaac and Karl have 4 choices each for their flights. So, for this scenario, there are also 4 * 4 * 4 = 64 outcomes.
c) If Karl is on a morning flight, Isaac and Jennifer again have 4 choices each for their flights. So, for this scenario, there are also 4 * 4 * 4 = 64 outcomes.
To find the total number of outcomes where exactly one of them has a morning flight, we add the number of outcomes for each scenario: 64 + 64 + 64 = 192 outcomes.