The problem states that there are three mutually exclusive outcomes with associated probabilities $p$, $\frac{p}{2}$, and $\frac{p}{4}$. Since these are probabilities of all possible outcomes, their sum must equal 1.

The total probability is given by: $p + \frac{p}{2} + \frac{p}{4} = 1$

To solve for $p$, first find a common denominator: $p + \frac{p}{2} + \frac{p}{4} = \frac{4p}{4} + \frac{2p}{4} + \frac{p}{4} = \frac{7p}{4}$

Now set this equal to 1: $\frac{7p}{4} = 1$

Multiply both sides by 4 to eliminate the fraction: $7p = 4$

Finally, solve for $p$: $p = \frac{4}{7}$

Thus, the value of $p$ is $\frac{4}{7}$.

Would you like any further details or clarification on this?

Here are some related questions for deeper understanding:

1. How are mutually exclusive outcomes defined in probability theory?
2. What happens if probabilities don't sum to 1 in a problem like this?
3. Can you explain why finding a common denominator is necessary in this problem?
4. How would you interpret the individual probabilities $\frac{p}{2}$ and $\frac{p}{4}$?
5. How would the solution change if there were four outcomes instead of three?

Tip: When dealing with probability distributions, always check that the sum of all probabilities equals 1 to ensure the solution is valid.