Six fair dice are thrown independently. The probability that three are exactly `2` different pairs (A pair is an ordered combination like `2,2,1,3,5,6`) is

Options

A. `5//72`
B. `26//72`
C. `125//144`
D. `5//36`

Explanation

To find the probability that when throwing six fair dice, there are exactly two different pairs, we can follow these steps:

### Step 1: Determine the Total Number of Outcomes
When throwing 6 fair dice, each die has 6 faces. Therefore, the total number of outcomes when throwing 6 dice is given by:

\[ \text{Total Outcomes} = 6^6 \]

### Step 2: Choose the Numbers for the Pairs
We need to select 2 different numbers to form the pairs. The total number of ways to choose 2 numbers from 6 is given by the combination formula:

\[ \text{Ways to choose 2 numbers} = \binom{6}{2} \]

### Step 3: Choose the Remaining Numbers
After choosing 2 numbers for the pairs, we need to select 2 more different numbers from the remaining 4 numbers. The number of ways to choose 2 numbers from these 4 is:

\[ \text{Ways to choose 2 from remaining 4} = \binom{4}{2} \]

### Step 4: Arrange the Dice
Now we need to arrange the 6 dice where we have 2 pairs (each appearing twice) and 2 other different numbers (each appearing once). The formula for arranging these dice is:

\[ \text{Ways to arrange} = \frac{6!}{2! \cdot 2!} \]

Where \(6!\) is the factorial of the total number of dice, and \(2!\) for each of the pairs.

### Step 5: Calculate the Required Probability
Now, we can combine all the parts to find the required probability:

\[ \text{Required Probability} = \frac{\binom{6}{2} \cdot \binom{4}{2} \cdot \frac{6!}{2! \cdot 2!}}{6^6} \]

### Step 6: Simplify the Expression
Now, we can compute the values:

1. \(\binom{6}{2} = 15\)
2. \(\binom{4}{2} = 6\)
3. \(6! = 720\)
4. \(2! = 2\)

Putting these values into the formula:

\[ \text{Required Probability} = \frac{15 \cdot 6 \cdot \frac{720}{2 \cdot 2}}{6^6} \]

Calculating the numerator:

\[ = \frac{15 \cdot 6 \cdot 180}{6^6} \]

Now, \(6^6 = 46656\):

\[ = \frac{16200}{46656} \]

### Step 7: Final Calculation
Now, we can simplify this fraction:

\[ = \frac{16200 \div 162}{46656 \div 162} = \frac{100}{288} \]

This can be simplified further:

\[ = \frac{25}{72} \]

Thus, the final answer for the probability that there are exactly two different pairs when throwing six fair dice is:

\[ \frac{25}{72} \]