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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.

A certain experiment has three possible outcomes. The outcomes are mutually exclusive and have probabilities p, p/2, and p/4, respectively. What is the value of p?

Learn how to solve a probability problem involving three mutually exclusive outcomes with given probabilities p, p/2, and p/4. This step-by-step guide shows how to solve for p using basic algebraic methods. Suitable for Grades 9-11.

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