Question

(a) A candy vending machine is out of order. The probability that you get a candy bar (with or without the return of your quarter) is \(\frac{1}{2}\), the probability that you get your quarter back (with or without candy) is \(\frac{1}{3}\), and the probability that you get both the candy and your money back is \(\frac{1}{12}\). What is the probability that you get nothing at all? Sugrestion: Sketch a geometric diagram similar to Figure \(3.1\), indicate regions representing the various possibilities and their probabilities; then set up a four-point samplc space and the associated probabilities of the points. (b) Suppose you. put another quarter into the candy vending machine of part (a). Set up the 16-point sample space corresponding to the possible results of your two attempts to bu? a candy bar, and find the probability that you get two candy bars (and no money back): that you get no candy and lose both quarters; that you just get your money back both times.

Short answer

The probability of getting nothing at all is \(\frac{1}{4}\). The probability of getting two candy bars and no quarters back is \(\frac{25}{144}\). The probability of losing both quarters and getting no candy is \(\frac{1}{16}\). The probability of getting your money back both times is \(\frac{1}{16}\).

Step-by-step solution

Identify the probabilities for single attempts

The probability of getting a candy bar is given as \(P(A) = \frac{1}{2}\). The probability of getting your quarter back is \(P(B) = \frac{1}{3}\). The probability of getting both a candy and your quarter back is \(P(A \cap B) = \frac{1}{12}\).

Use the Inclusion-Exclusion Principle to find the probability of getting nothing

The probability of getting either a candy or your quarter back or both is \(P(A \cup B) = P(A) + P(B) - P(A \cap B) = \frac{1}{2} + \frac{1}{3} - \frac{1}{12} = \frac{6}{12} + \frac{4}{12} - \frac{1}{12} = \frac{9}{12} = \frac{3}{4}\). Therefore, the probability of getting nothing at all is \(P(\text{Nothing}) = 1 - P(A \cup B) = 1 - \frac{3}{4} = \frac{1}{4}\).

Set up the sample space for two attempts

Let \(C_1\) and \(Q_1\) represent getting candy and quarter back respectively in the first attempt, and \(C_2\) and \(Q_2\) represent them in the second attempt. The sample space for two attempts is \(\{ (C_1Q_1, C_2Q_2), (C_1Q_1, C_2 \overline{Q}_2), (C_1Q_1, \overline{C}_2Q_2), (C_1 \overline{Q}_1, C_2Q_2), (C_1 \overline{Q}_1, C_2 \overline{Q}_2), ... , (\overline{C}_1 \overline{Q}_1, \overline{C}_2 \overline{Q}_2) \}\), where \(\overline{C}\) and \(\overline{Q}\) denote not getting candy and not getting quarter back respectively.

Calculate specific probabilities for two attempts

The probability of getting two candy bars and no quarters back is \(P(C_1 \overline{Q}_1, C_2 \overline{Q}_2) = P(C_1 \overline{Q}_1) \times P(C_2 \overline{Q}_2) = (P(C_1) - P(C_1Q_1))(P(C_2) - P(C_2Q_2)) = (\frac{1}{2} - \frac{1}{12})(\frac{1}{2} - \frac{1}{12}) = (\frac{5}{12})(\frac{5}{12}) = \frac{25}{144}\). The probability of getting no candy and losing both quarters is \(P(\overline{C}_1 \overline{Q}_1, \overline{C}_2 \overline{Q}_2) = P(\overline{C}_1 \overline{Q}_1) \times P(\overline{C}_2 \overline{Q}_2) = (1 - P(A \cup B))(1 - P(A \cup B)) = (\frac{1}{4})(\frac{1}{4}) = \frac{1}{16}\). The probability of just getting your money back both times is \(P(\overline{C}_1Q_1, \overline{C}_2Q_2) = P(\overline{C}_1Q_1) \times P(\overline{C}_2Q_2) = (P(Q_1) - P(C_1Q_1))(P(Q_2) - P(C_2Q_2)) = (\frac{1}{3} - \frac{1}{12})(\frac{1}{3} - \frac{1}{12}) = (\frac{1}{4})(\frac{1}{4}) = \frac{1}{16}\).

Key concepts

Sample Space

A sample space is the set of all possible outcomes in a probability experiment. When dealing with more complicated scenarios, such as the candy vending machine problem, listing all outcomes systematically helps us understand the problem better.

For example, in part (a) of the problem, we analyze the outcomes of a single attempt to get a candy bar or get your quarter back. Here, our sample space might include outcomes such as getting only candy, only your quarter, both, or neither.

In part (b) of the problem, things become a bit more complex as we deal with two attempts. The sample space here consists of all combinations of the outcomes from these two attempts. By systematically listing these combinations, we can more accurately calculate the probabilities of various events.

Inclusion-Exclusion Principle

The inclusion-exclusion principle helps us find the probability of the union of two events by considering each event's probability and subtracting the overlap.

Here's how we implemented it in the candy machine problem for part (a):

• First, we identified the probabilities of getting candy (\(P(A) = \frac{1}{2}\)), getting your quarter back (\(P(B) = \frac{1}{3}\)), and both (\(P(A \cap B) = \frac{1}{12}\)).
• Then, we calculated the probability of getting either candy or your quarter back (or both) using the formula: \(P(A \cup B) = P(A) + P(B) - P(A \cap B)\). This simplifies to: \(\frac{1}{2} + \frac{1}{3} - \frac{1}{12} = \frac{3}{4}\).

In the end, the probability of getting nothing at all is the complement of this probability: \(1 - P(A \cup B) = \frac{1}{4}\).

Conditional Probability

Conditional probability measures the likelihood of an event occurring, given that another event has already occurred. It's denoted by \(P(A|B)\), which reads as 'the probability of A given B'.

While the main problem did not directly require conditional probability, understanding this concept is still vital. For instance, if you knew you got your quarter back, what is the probability that you also got candy? You would use the formula:
\(\frac{P(A \cap B)}{P(B)} = \frac{1/12}{1/3} = \frac{1}{4}\)
This means that given the information of getting your quarter back, there’s a 25% chance you'll also get the candy.

Independent Events

Independent events are events where the occurrence of one doesn't affect the probability of the other.

Determining whether events are independent or not helps simplify probability calculations greatly.
For example, if we consider each attempt to buy a candy bar as an independent event, the overall probability of outcomes across two attempts is a product of the probabilities from each attempt. For instance:

• The probability of getting two candy bars and no money back is calculated as: \(P(C_1 \overline{Q}_1, C_2 \overline{Q}_2) = P(C_1 \overline{Q}_1) \times P(C_2 \overline{Q}_2) = (\frac{5}{12}) \times (\frac{5}{12}) = \frac{25}{144}\).
• The probability of losing both quarters and getting no candy is: \(P(\overline{C}_1 \overline{Q}_1, \overline{C}_2 \overline{Q}_2) = (\frac{1}{4}) \times (\frac{1}{4}) = \frac{1}{16}\).