If an unbiased die, marked with -2, -1, 0, 1, 2, 3 on its faces, is thrown five times, then the prob

Q: Wait, why can't we just use $1 - P(\text{negative})$? That seems much faster.

That is a classic trap! If you calculate $1 - P(\text{negative})$, you are only subtracting the cases where the product is negative. However, you are forgetting the case where the product is $0$. If you roll a $0$, the product is neither positive nor negative. By using $1 - P(\text{negative})$, you would accidentally include the cases where the product is $0$ in your 'positive' result. Always account for the zero!

Q: Why is the probability of getting a positive number $1/2$ and a negative number $1/3$?

Great question. We are looking at the sample space of the die, which is $\lbrace -2, -1, 0, 1, 2, 3 \rbrace$. There are $6$ total faces. The positive faces are $\lbrace 1, 2, 3 \rbrace$, so there are $3$ favorable outcomes. Thus, $P(\text{positive}) = \frac{3}{6} = \frac{1}{2}$. The negative faces are $\lbrace -2, -1 \rbrace$, so there are $2$ favorable outcomes. Thus, $P(\text{negative}) = \frac{2}{6} = \frac{1}{3}$. We ignore the $0$ face because it ruins our product.

Q: Why do we use the Binomial Distribution here?

We use the Binomial Distribution because we have a fixed number of trials ($n=5$), each trial is independent (the die doesn't remember the previous roll), and each trial has only two outcomes of interest for our specific case: either we get a negative number (our 'success' for the binomial count) or we get a positive number. Since we are excluding $0$, we are effectively working in a reduced sample space of $5$ faces.

Question

If an unbiased die, marked with $-2, -1, 0, 1, 2, 3$ on its faces, is thrown five times, then the probability that the product of the outcomes is positive, is

Options:

(A) $\frac{881}{2592}$

(B) $\frac{521}{2592}$

(C) $\frac{440}{2592}$

(D) $\frac{27}{288}$

Sigma Percentile · Accepted Answer

Welcome, future engineers. Today, we are going to dismantle a probability problem that is designed to test not just your math, but your ability to stay calm under pressure.

Imagine standing before a die that is not your standard $1$ to $6$ cube. This die is marked with $\lbrace -2, -1, 0, 1, 2, 3 brace$. You are tasked with throwing it five times and ensuring the product of those five outcomes is strictly positive.

### The Anatomy of the Trap

The first thing you must do is look at the faces. The presence of $0$ is the most dangerous element here. In the world of products, $0$ is a black hole.

If you roll a $0$ even once, the entire product collapses to $0$. Since the problem demands a strictly positive product, we must treat $0$ as a forbidden outcome.

We are effectively working with a reduced sample space of five faces: $\lbrace -2, -1, 1, 2, 3 brace$. Within this set, we have two types of numbers: positive numbers $\lbrace 1, 2, 3 brace$ and negative numbers $\lbrace -2, -1 brace$.

The probability of rolling a positive number is $p = \frac{3}{6} = \frac{1}{2}$, and the probability of rolling a negative number is $q = \frac{2}{6} = \frac{1}{3}$.

### The Parity Logic

Now, how do we ensure a positive product? We know that a positive number multiplied by any number of positive numbers remains positive. The real action happens with the negative numbers.

A negative times a negative is a positive. Therefore, to keep our final product positive, we must have an even number of negative outcomes.

In five throws, we can have $0$ negative outcomes, $2$ negative outcomes, or $4$ negative outcomes. We cannot have $6$ because we only have $5$ throws. This is where the Binomial Distribution becomes our best friend.

### The Three Cases

We must calculate the probability for each of these three scenarios independently.

**Case 1: Zero negative outcomes.** 
This means all $5$ throws must be positive. The probability is:

$$P_0 = \binom{5}{0} 	imes \left(\frac{1}{2}ight)^5 	imes \left(\frac{1}{3}ight)^0 = 1 	imes \frac{1}{32} 	imes 1 = \frac{1}{32}$$

**Case 2: Two negative outcomes.** 
This means $2$ throws are negative and $3$ are positive. The probability is:

$$P_2 = \binom{5}{2} 	imes \left(\frac{1}{2}ight)^3 	imes \left(\frac{1}{3}ight)^2 = 10 	imes \frac{1}{8} 	imes \frac{1}{9} = \frac{10}{72} = \frac{5}{36}$$

**Case 3: Four negative outcomes.** 
This means $4$ throws are negative and $1$ is positive. The probability is:

$$P_4 = \binom{5}{4} 	imes \left(\frac{1}{2}ight)^1 	imes \left(\frac{1}{3}ight)^4 = 5 	imes \frac{1}{2} 	imes \frac{1}{81} = \frac{5}{162}$$

### The Final Synthesis

Now, we bring it all together. Since these cases are mutually exclusive—you cannot have both $0$ negatives and $2$ negatives at the same time—we sum them up.

The total probability is:

$$P_{total} = \frac{1}{32} + \frac{5}{36} + \frac{5}{162}$$

Finding the least common multiple of $32, 36,$ and $162$ gives us $2592$. Converting the fractions, we get:

$$P_{total} = \frac{81}{2592} + \frac{360}{2592} + \frac{80}{2592}$$

Adding these numerators, $81 + 360 + 80$, yields $521$. Thus, our final probability is **$\frac{521}{2592}$**.

Animated Solution for Mathematics - Probability: If an unbiased die, marked with $- 2, - 1, 0, 1, 2, 3$ on its faces, is thrown five times, then the probability that the product of the outcomes is positive, is

Select Answer:

Visualized Solution

Understanding the Die Faces

Condition for a Positive Product

Probability of Individual Outcomes

Valid Combinations for $5$ Throws

Case 1: Five Positive Outcomes

Case 2: Three Positive, Two Negative

Case 3: One Positive, Four Negative

Total Probability

Final Calculation

The Sigma Insight: Binomial Distribution

The Anatomy of the Trap

The Parity Logic

The Three Cases

The Final Synthesis

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