Question

A bag contains 5 balls of unknown colors. There are equal chances that out of these five balls, there may be 0 or 1 or 2 or 3 or 4 or 5 red balls. A ball is taken out from the bag at random and is found to be red. The probability that it is the only red ball in the bag is:

• A. $ \frac{1}{5} $
• B. $ \frac{1}{6} $
• C. $ \frac{1}{15} $
• D. $ \frac{1}{30} $

Hint:

When solving problems involving conditional probabilities, use Bayes' theorem to relate the conditional probability to the joint and marginal probabilities. Ensure all cases are considered systematically.

Answer 1

The Correct Option is C

Step 1: Define the problem setup.
There are 6 possible scenarios for the number of red balls in the bag:
0 red balls,
1 red ball,
2 red balls,
3 red balls,
4 red balls,
5 red balls.
Each scenario is equally likely, so the probability of each scenario is: $$ P(\text{Scenario}) = \frac{1}{6}. $$ Let \( R \) denote the event that a randomly drawn ball is red. We need to find the conditional probability: $$ P(1 \text{ red ball} \mid R). $$ Step 2: Use Bayes' theorem.
By Bayes' theorem:
$$ P(1 \text{ red ball} \mid R) = \frac{P(R \mid 1 \text{ red ball}) \cdot P(1 \text{ red ball})}{P(R)}. $$ Step 3: Compute \( P(R \mid 1 \text{ red ball}) \).
If there is exactly 1 red ball, the probability of drawing a red ball is: $$ P(R \mid 1 \text{ red ball}) = \frac{1}{5}. $$ Step 4: Compute \( P(R) \).
The total probability of drawing a red ball is the sum of the probabilities of drawing a red ball under each scenario, weighted by the probability of each scenario: $$ P(R) = \sum_{k=0}^{5} P(R \mid k \text{ red balls}) \cdot P(k \text{ red balls}), $$ where \( k \) is the number of red balls. For each scenario:
If \( k = 0 \), \( P(R \mid k = 0) = 0 \),
If \( k = 1 \), \( P(R \mid k = 1) = \frac{1}{5} \),
If \( k = 2 \), \( P(R \mid k = 2) = \frac{2}{5} \),
If \( k = 3 \), \( P(R \mid k = 3) = \frac{3}{5} \),
If \( k = 4 \), \( P(R \mid k = 4) = \frac{4}{5} \),
If \( k = 5 \), \( P(R \mid k = 5) = 1 \).
Thus: $$ P(R) = \sum_{k=0}^{5} P(R \mid k \text{ red balls}) \cdot P(k \text{ red balls}) = \sum_{k=0}^{5} \frac{k}{5} \cdot \frac{1}{6}. $$ Simplify:
$$ P(R) = \frac{1}{6} \left( 0 + \frac{1}{5} + \frac{2}{5} + \frac{3}{5} + \frac{4}{5} + 1 \right) = \frac{1}{6} \left( \frac{0 + 1 + 2 + 3 + 4 + 5}{5} \right) = \frac{1}{6} \cdot \frac{15}{5} = \frac{1}{6} \cdot 3 = \frac{1}{2}. $$ Step 5: Compute \( P(1 \text{ red ball} \mid R) \).
Substitute into Bayes' theorem:
$$ P(1 \text{ red ball} \mid R) = \frac{P(R \mid 1 \text{ red ball}) \cdot P(1 \text{ red ball})}{P(R)} = \frac{\frac{1}{5} \cdot \frac{1}{6}}{\frac{1}{2}} = \frac{\frac{1}{30}}{\frac{1}{2}} = \frac{1}{30} \cdot 2 = \frac{1}{15}. $$ Step 6: Final Answer.
$$ \boxed{\frac{1}{15}}. $$