Given:
- Coefficient of concurrent deviation = 1/√3
- Number of concurrent deviations = 6

To find:
- Value of p (number of pairs of observations)

Solution:
The coefficient of concurrent deviation measures the average variability between two sets of observations. It is defined as the square root of the ratio of the mean square deviation of one set to the mean square deviation of the other set.

In this case, the coefficient of concurrent deviation is given as 1/√3. Let's denote the mean square deviation of one set as x and the mean square deviation of the other set as y.

So, we have:
Coefficient of concurrent deviation = √(x/y) = 1/√3

Squaring both sides of the equation, we get:
x/y = (1/√3)^2
x/y = 1/3

Now, the number of concurrent deviations is given as 6. Since each concurrent deviation involves a pair of observations, we can say that there are 6/2 = 3 pairs of observations.

We can calculate the sum of the squares of the deviations for each pair of observations using the formula:
Sum of squares of deviations = (x + y)

Substituting the value of x/y from the previous equation, we have:
Sum of squares of deviations = (1/3 + 1) = (4/3)

Since we have p pairs of observations, we can write the equation as:
p * (4/3) = 6

Simplifying the equation, we get:
p = (6 * 3) / 4
p = 18 / 4
p = 4.5

Since the number of pairs of observations cannot be a decimal value, we can conclude that it is not possible to have 4.5 pairs of observations. Therefore, there is no value of p that satisfies the given conditions.

Conclusion:
The value of p (number of pairs of observations) cannot be determined based on the given information.