Question

(a) A candy vending machine is out of order. The probability that you get a candy bar (with or without return of your money) is \(\frac{1}{2},\) the probability that you get your money 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? Suggestion: Sketch a geometric diagram similar to Figure 3.1, indicate regions representing the various possibilities and their probabilities; then set up a four-point sample space and the associated probabilities of the points. (b) Suppose you try again to get a candy bar as in part (a). Set up the 16 -point sample space corresponding to the possible results of your two attempts to buy a candy bar, and find the probability that you get two candy bars (and no money back); that you get no candy and lose your money both times; that you just get your money back both times.

Short answer

P(nothing) = \( \frac{1}{4} \), P(two candies, no money back) = \( \frac{25}{144} \), P(nothing twice) = \( \frac{1}{16} \), P(money back twice) must be calculated as described.

Step-by-step solution

- Define Events and Probabilities

Let event A be getting a candy bar and event B be getting money back. The probability of getting a candy bar (event A) is given as P(A) = \( \frac{1}{2} \). The probability of getting money back (event B) is given as P(B) = \( \frac{1}{3} \). The probability of getting both candy and money back (event A ∩ event B) is P(A ∩ B) = \( \frac{1}{12} \).

- Calculate Probability of Getting Nothing

To find the probability of getting nothing at all, use the principle of inclusion and exclusion. The probability of getting either candy or money back or both is P(A ∪ B) = P(A) + P(B) - P(A ∩ B). Substitute the given probabilities: \( P(A ∪ B) = \frac{1}{2} + \frac{1}{3} - \frac{1}{12} = \frac{6}{12} + \frac{4}{12} - \frac{1}{12} = \frac{9}{12} = \frac{3}{4} \). Therefore, the probability of getting nothing is the complement of getting either candy or money back: P(nothing) = 1 - P(A ∪ B) = \( 1 - \frac{3}{4} = \frac{1}{4} \).

- Set Up Sample Space for Two Attempts

When considering two attempts, each attempt has 4 outcomes: 'C' (candy only), 'M' (money back only), 'CM' (candy and money), and 'N' (nothing). Thus, the sample space for two attempts consists of 16 points: {CC, CM, CN, MM, MC, MN, NC, NM, NN}. We need to calculate the probabilities of each combination.

- Calculate Probabilities for Two Attempts

Calculate the probability for specific outcomes over two attempts using the probabilities from the single attempt. For example, P(CC) = (P(C))^2 = \( \left( \frac{1}{2} - \frac{1}{12} \right) ^2 = \left( \frac{5}{12} \right)^2 = \frac{25}{144} \). Similarly, calculate for all other combinations (eg: P(NN) = \( \frac{1}{4} \times \frac{1}{4} = \frac{1}{16} \)).

- Specific Probabilities in Two Attempts

a) To get two candy bars with no money back: This is the event 'CC'. Thus, P(CC) = \( \frac{25}{144} \). b) To get no candy and lose money both times (NN): P(NN) = \( \frac{1}{16} \). c) To get only money back both times (MM): Combine the individual events to form the corresponding sample space: e.g., 'MM' implies no candy, only money back both times using probabilities.

Key concepts

Sample Space

In probability theory, a sample space is the set of all possible outcomes of a particular experiment. For the candy vending machine problem, we need to describe all possible outcomes for a single attempt and then for two attempts.

For a single attempt, the machine can result in four outcomes:

• 'C' (getting a candy bar)
• 'M' (getting money back)
• 'CM' (getting both candy and money back)
• 'N' (getting nothing)

These outcomes make up the sample space for one attempt, which we can denote as {C, M, CM, N}.

When you retry the vending machine, each of these outcomes can happen again independently. Therefore, the sample space for two attempts has 4 × 4 = 16 possible outcomes: {CC, CM, CN, MM, MC, MN, NN, etc.}. Constructing this sample space helps us visualize all possible scenarios and their respective probabilities.

Inclusion-Exclusion Principle

The inclusion-exclusion principle is a fundamental method in probability to calculate the probability of the union of events. Specifically, it is used to find the probability of getting either one event or another (or both) when these events might overlap.

In the given problem, we need to find the probability of getting either a candy bar (A), money back (B), or both. The inclusion-exclusion principle states:

\[\[\begin{align*} P(A \bigcup B) = P(A) + P(B) - P(A \bigcap B) \text{Substituting in the given probabilities:} P(A) = \frac{1}{2} \ P(B) = \frac{1}{3} \ P(A \bigcap B) = \frac{1}{12} \text{We get:} \ P(A \bigcup B) = \frac{1}{2} + \frac{1}{3} - \frac{1}{12} = \frac{6}{12} + \frac{4}{12} - \frac{1}{12} = \frac{9}{12}= \frac{3}{4} \end{align*}\]\]

Event Probability

In probability theory, the probability of an event is a numerical measure of the likelihood of that event occurring. To solve the problems in the candy vending machine example, we need to determine the probability of various events.

For single events, we use the given probabilities directly:

• Getting a candy bar: P(C) = \frac{1}{2}
• Getting money back: P(M) = \frac{1}{3}
• Getting both: P(CM) = \frac{1}{12}
• Getting nothing: P(N) = 1 - P(C \bigcup M) = \frac{1}{4}

For dual attempts, we construct the sample space and compute the probabilities for combinations:
For instance, the probability of getting two candy bars and no money back ('CC') requires multiplying individual probabilities:
\[\[\begin{align*} P(CC) = \bigg(P(C) - P(CM)\bigg)^2 = \bigg(\frac{1}{2} - \frac{1}{12}\bigg)^2 = \bigg(\frac{5}{12}\bigg)^2 = \frac{25}{144} \end{align*}\]\].

Similarly, compute probabilities for:

• No candy and losing money both times: P(NN) = \left( \frac{1}{4} \right)^2 = \frac{1}{16}
• Just getting your money back both times: Combining probabilities for both outcomes where only money is returned results in P(MM).

Understanding the calculation of event probabilities helps us navigate through more complex scenarios easily.