If S = {a,b,c} with P(a) = 2P(b) = 4P(c), find P(a)

We are given the set \( S = \{a, b, c\} \) and the probabilities such that:

\[
P(a) = 2P(b) = 4P(c)
\]

This means:

\[
P(a) = 2P(b) \quad \text{and} \quad P(a) = 4P(c)
\]

The total probability for any set must sum to 1:

\[
P(a) + P(b) + P(c) = 1
\]

Now, express \( P(b) \) and \( P(c) \) in terms of \( P(a) \):

1. \( P(a) = 2P(b) \) implies \( P(b) = \frac{P(a)}{2} \)
2. \( P(a) = 4P(c) \) implies \( P(c) = \frac{P(a)}{4} \)

Substitute these into the total probability equation:

\[
P(a) + \frac{P(a)}{2} + \frac{P(a)}{4} = 1
\]

Find a common denominator:

\[
P(a) \left( 1 + \frac{1}{2} + \frac{1}{4} \right) = 1
\]

Simplify the terms in parentheses:

\[
1 + \frac{1}{2} + \frac{1}{4} = \frac{4}{4} + \frac{2}{4} + \frac{1}{4} = \frac{7}{4}
\]

So the equation becomes:

\[
P(a) \cdot \frac{7}{4} = 1
\]

Solve for \( P(a) \):

\[
P(a) = \frac{4}{7}
\]

Thus, the probability of \( a \) is \( P(a) = \frac{4}{7} \).

Would you like a further breakdown or more details on any step?

Here are 5 related questions you might find interesting:
1. How do you calculate probabilities for a larger set?
2. What happens if probabilities aren't proportional, but follow another relationship?
3. Can this method be applied to continuous probability distributions?
4. How would the solution change if there were more than 3 events?
5. How can this type of problem appear in real-life scenarios like gambling or statistics?

**Tip:** Always make sure the total probability sums to 1 when dealing with discrete events.

Discover how to find the probability of event a when given proportional relationships between probabilities of events in a set. This step-by-step solution covers key probability concepts, equations, and the Law of Total Probability. Suitable for students in Grades 10-12.

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