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. `(440)/(2592)`
B. `(81)/(2592)`
C. `(27)/(282)`
D. `(521)/(2592)`

Answer

To find the probability that the product of the outcomes when throwing an unbiased die marked with `-2, -1, 0, 1, 2, 3` five times is positive, we can follow these steps:

### Step 1: Identify the outcomes
The die has the following outcomes:
- Negative outcomes: `-2, -1` (2 outcomes)
- Zero outcome: `0` (1 outcome)
- Positive outcomes: `1, 2, 3` (3 outcomes)

### Step 2: Determine conditions for a positive product
For the product of the outcomes to be positive, we need to consider the following cases:
1. All outcomes are positive.
2. An odd number of negative outcomes (1 or 3) and the rest positive (ensuring there are no zeros).

### Step 3: Calculate the total number of outcomes
When the die is thrown 5 times, the total number of outcomes is:
\[ 6^5 = 7776 \]

### Step 4: Calculate the probability for each case

#### Case 1: All outcomes are positive
The probability of getting a positive outcome in one throw is:
\[ P(\text{positive}) = \frac{3}{6} = \frac{1}{2} \]
Thus, for 5 throws:
\[ P(\text{all positive}) = \left(\frac{1}{2}\right)^5 = \frac{1}{32} \]

#### Case 2: 3 positive and 2 negative
We need to choose 3 positive outcomes from 5 throws, and the remaining 2 must be negative. The number of ways to choose 2 negative outcomes from 5 throws is given by:
\[ \binom{5}{2} = 10 \]
The probability of getting 3 positive and 2 negative outcomes is:
\[ P(3 \text{ positive, } 2 \text{ negative}) = \binom{5}{2} \left(\frac{3}{6}\right)^3 \left(\frac{2}{6}\right)^2 = 10 \cdot \left(\frac{1}{2}\right)^3 \cdot \left(\frac{1}{3}\right)^2 = 10 \cdot \frac{1}{8} \cdot \frac{1}{9} = \frac{10}{72} = \frac{5}{36} \]

#### Case 3: 1 positive and 4 negative
The number of ways to choose 1 positive outcome from 5 throws is:
\[ \binom{5}{1} = 5 \]
The probability of getting 1 positive and 4 negative outcomes is:
\[ P(1 \text{ positive, } 4 \text{ negative}) = \binom{5}{1} \left(\frac{3}{6}\right)^1 \left(\frac{2}{6}\right)^4 = 5 \cdot \left(\frac{1}{2}\right) \cdot \left(\frac{1}{3}\right)^4 = 5 \cdot \frac{1}{2} \cdot \frac{1}{81} = \frac{5}{162} \]

### Step 5: Combine the probabilities
Now, we combine the probabilities from all cases:
\[ P(\text{positive product}) = P(\text{all positive}) + P(3 \text{ positive, } 2 \text{ negative}) + P(1 \text{ positive, } 4 \text{ negative}) \]
\[ = \frac{1}{32} + \frac{5}{36} + \frac{5}{162} \]

### Step 6: Find a common denominator
The least common multiple of 32, 36, and 162 is 1296. We convert each fraction:
\[ \frac{1}{32} = \frac{1 \cdot 40.5}{32 \cdot 40.5} = \frac{40.5}{1296} \]
\[ \frac{5}{36} = \frac{5 \cdot 36}{36 \cdot 36} = \frac{180}{1296} \]
\[ \frac{5}{162} = \frac{5 \cdot 8}{162 \cdot 8} = \frac{40}{1296} \]

### Step 7: Add the fractions
Now we add them:
\[ P(\text{positive product}) = \frac{40.5 + 180 + 40}{1296} = \frac{260.5}{1296} \]

### Final Result
Thus, the probability that the product of the outcomes is positive is:
\[ \frac{521}{2592} \]