Question

For the experiments, list the simple events in the sample space, assign probabilities to the simple events, and find the required probabilities. A bowl contains five candies-red, brown, yellow, blue, and orange. Draw two candies at random, one for you to eat, and one for a friend. What is the probability that you get the orange candy and your friend does not get the red one?

Short answer

Answer: The probability is 3/20.

Step-by-step solution

List the sample space

In this case, we have five different candies and are drawing two candies at random, one for ourselves and one for our friend. To find the sample space, we can list all the possible pairs of candies that we can get. We will use the first letter of each color as an abbreviation for the candy, so for example, R stands for red, B for blue, etc. There are 20 possible pairs: 1. (R, R) 2. (R, B) 3. (R, Y) 4. (R, O) 5. (R, B) 6. (B, R) 7. (B, Y) 8. (B, O) 9. (B, B) 10. (B, Y) 11. (Y, R) 12. (Y, B) 13. (Y, O) 14. (Y, Y) 15. (Y, B) 16. (O, R) 17. (O, B) 18. (O, Y) 19. (O, O) 20. (O, B)

Assign probabilities to the sample space

Since there are five candies, and we are drawing two of them, there are 5 choose 2 = 10 unique combinations, with each combination being equally likely. Including the combinations when the two candies are of the same color, there are 20 pairs of candies that can be drawn. Therefore, each simple event in the sample space has a probability of 1/20. Now we assign the probability 1/20 to all the pairs listed in step 1.

Calculate the probability of the event of interest

We are interested in the probability that we get the orange candy (O), and our friend does not get the red candy (R). This implies that we are looking for the probability of getting the following pairs: (O, B), (O, Y), or (O, B). Since each of these pairs has a probability of 1/20, we can calculate the overall probability by summing their probabilities: P(we get O and friend does not get R) = P(O, B) + P(O, Y) + P(O, B) = \frac{1}{20} + \frac{1}{20} + \frac{1}{20} = \frac{3}{20} Hence, the probability that we get the orange candy while our friend does not get the red one is 3/20.

Key concepts

Sample Space

In probability, the sample space is a crucial concept. It represents all possible outcomes of a particular experiment. When dealing with probability problems, identifying the sample space is often the first step. In our candy bowl scenario, each outcome consists of picking two candies: one for you and another for your friend.
You can think of the sample space as a list of all pairs available. For the given problem, there are five candies - red, brown, yellow, blue, and orange. When two candies are drawn, simply consider all combinations. If the first candy is the one you pick, the pairs denote the candy you and your friend get together.
Here’s a friendly reminder: remember that each candy can be paired with itself or a different candy. This means the sample space isn't just the number of candies, but every unique way two candies can be distributed between you and your friend.

• Think first about swapping roles; getting a yellow then a red is different from a red then yellow due to assignment roles.
• Ensure you don't miss pairs by considering each candy selection methodically.

Combinatorics

Combinatorics is the area of mathematics that studies the counting, arrangement, and combination of objects. In our exercise, it is essentially used to understand how candies can be distributed between you and a friend.
Calculating combinatory arrangements involves using formulas to determine different ways an event can occur. For drawing two candies out of five, you use combinations to ensure you include every viable way two candies can pair without regard to the order of draw. In simple terms, you are interested in selecting, not arranging, the candies.
The formula applied is 5 choose 2, representing the ways to choose 2 items from a set of 5. Mathematically, it's denoted as \( \binom{5}{2} \). But here, also consider the permutations where the order does matter, which ultimately means recalculating for all circumstances.

• Understand that combinatorics helps simplify determining the complete range of possible outcomes.
• Make a distinction between permutations (order matters) and combinations (order does not matter).

Event Probability

Event probability calculates the likelihood that a particular outcome will occur. After knowing all possible outcomes (the sample space) and understanding the combinations (combinatorics), you can assign probabilities to each specific event.
Every pair (event) in our sample space of candy choices is assumed equally likely, with a total of 20 pairs possible. Therefore, each pair has the probability of \( \frac{1}{20} \). Here, our aim is to find the probability of a favourable condition – receiving the orange candy while the friend doesn't receive red. This specific condition narrows down to three preferable pairs: \((O, B), (O, Y), (O, B)\).
Combining probabilities of individual events that form the condition you’re interested in helps you find your answer, which is simply adding up these individual probabilities. For this exercise, the answer is 3 out of 20: \( \frac{3}{20} \).

• Ensure clarity by isolating the conditions and focusing on what's needed to satisfy them.
• Add values when similar desirable outcomes exist in probability calculations.