The problem describes an experiment with three possible mutually exclusive outcomes, meaning the total probability for all outcomes must sum to 1. The probabilities of the outcomes are $p$, $\frac{p}{2}$, and $\frac{p}{4}$.

To find the value of $p$, we add the three probabilities and set them equal to 1:

$p + \frac{p}{2} + \frac{p}{4} = 1$

Step 1: Combine the terms

To combine the terms on the left-hand side, express all terms with a common denominator. The least common denominator of 1, 2, and 4 is 4. Rewriting each term with denominator 4:

$p = \frac{4p}{4}, \quad \frac{p}{2} = \frac{2p}{4}, \quad \frac{p}{4} = \frac{p}{4}$

Now add these fractions:

$\frac{4p}{4} + \frac{2p}{4} + \frac{p}{4} = \frac{7p}{4}$

Step 2: Solve for $p$

We now have the equation:

$\frac{7p}{4} = 1$

Multiply both sides by 4 to eliminate the denominator:

$7p = 4$

Finally, divide by 7:

$p = \frac{4}{7}$

Conclusion:

The value of $p$ is $\frac{4}{7}$.

Would you like further details or have any other questions?

Here are 5 related questions you might find useful:

1. How do mutually exclusive events affect probability distributions?
2. How would you solve if there were more than three possible outcomes?
3. Can probabilities ever exceed 1 or be negative?
4. How do you check if a given set of probabilities is valid for a probability distribution?
5. What are some applications of probability in real-world experiments?

Tip: When dealing with probabilities, always ensure that the sum of all possible outcomes equals 1 to validate your distribution.