Question

Let S={\(a,b,c\)} be the sample space with the associated probabilities satisfying \(P(a)=2P(b)\) and \(P(b)=2P(c).\)Then the value of \(P(a)\) is 

• A. \(\dfrac{1}{5}\)
• B. \(\dfrac{2}{7}\)
• C. \(\dfrac{4}{7}\)
• D. \(\dfrac{1}{6}\)
• E. \(\dfrac{1}{7}\)

Answer 1

The Correct Option is C

Given:

• Sample space \( S = \{a, b, c\} \)
• \( P(a) = 2P(b) \)
• \( P(b) = 2P(c) \)

 

We need to find the value of \( P(a) \).

Step 1: Express all probabilities in terms of \( P(c) \): \[ P(b) = 2P(c) \] \[ P(a) = 2P(b) = 2 \times 2P(c) = 4P(c) \]

Step 2: Apply the total probability rule: \[ P(a) + P(b) + P(c) = 1 \] \[ 4P(c) + 2P(c) + P(c) = 1 \] \[ 7P(c) = 1 \implies P(c) = \frac{1}{7} \]

Step 3: Calculate \( P(a) \): \[ P(a) = 4P(c) = 4 \times \frac{1}{7} = \frac{4}{7} \]

The correct answer is (C) \( \frac{4}{7} \).

Answer 2

Let \( S = \{a, b, c\} \) be the sample space.

The probabilities associated with the sample space satisfy:

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

Since \( S \) is the sample space, the sum of probabilities of all events in \( S \) must be equal to 1. Therefore,

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

We have \( P(a) = 2P(b) \) and \( P(b) = 2P(c) \).

Substituting \( P(b) = 2P(c) \) into \( P(a) = 2P(b) \), we get:

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

Now we substitute \( P(a) = 4P(c) \) and \( P(b) = 2P(c) \) into the equation \( P(a) + P(b) + P(c) = 1 \):

\[ 4P(c) + 2P(c) + P(c) = 1 \] \[ 7P(c) = 1 \] \[ P(c) = \frac{1}{7}. \]

Now we can find \( P(b) \) and \( P(a) \):

\[ P(b) = 2P(c) = 2\left(\frac{1}{7}\right) = \frac{2}{7} \] \[ P(a) = 4P(c) = 4\left(\frac{1}{7}\right) = \frac{4}{7}. \]

Therefore, \( P(a) = \frac{4}{7} \), \( P(b) = \frac{2}{7} \), and \( P(c) = \frac{1}{7} \).

We can check that these probabilities sum to 1:

\[ P(a) + P(b) + P(c) = \frac{4}{7} + \frac{2}{7} + \frac{1}{7} = \frac{7}{7} = 1. \]

The value of \( P(a) \) is \( \frac{4}{7} \).