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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?

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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.

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 for the probability p in an experiment with three mutually exclusive outcomes. This problem uses algebra and the law of total probability to combine fractional probabilities and find p. Suitable for students in Grades 8-10.

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