Question

A bag contains 6 balls. If three balls are drawn at a time and all of them are found to be green, then the probability that exactly 5 of the balls in the bag are green is:

• A. \( \frac{4}{35} \)
• B. \( \frac{5}{35} \)
• C. \( \frac{2}{7} \)
• D. \( \frac{1}{7} \)

Hint:

For probability problems involving conditional cases, Bayes' Theorem helps in determining probabilities given prior conditions.

Answer 1

The Correct Option is C

Step 1: Understanding the Problem We are tasked with finding the probability that exactly 5 of the balls in the bag are green given that 3 randomly drawn balls are all green.
Step 2: Applying Bayes' Theorem Let the events be defined as follows: - \( A \) = "Exactly 5 of the 6 balls are green." - \( B \) = "Three balls drawn are all green." We need to calculate \( P(A | B) \), the probability that there are exactly 5 green balls given that 3 drawn balls are green. By Bayes' Theorem: \[ P(A | B) = \frac{P(B | A) \cdot P(A)}{P(B)} \]
Step 3: Calculating Each Probability
Step 3a: Probability of \( A \) (Probability of exactly 5 green balls in the bag) Since 5 out of 6 balls are green: \[ P(A) = \frac{1}{3} \quad \text{(Assuming equally likely cases: 5 green or 6 green)} \]
Step 3b: Probability of \( B | A \) (Probability of drawing 3 green balls if there are exactly 5 green balls) If there are 5 green balls in the bag, the total ways to choose 3 balls that are all green: - Choose 3 balls from 5 green balls: \( \binom{5}{3} = 10 \) - Choose any 3 balls out of 6 total balls: \( \binom{6}{3} = 20 \) Thus, \[ P(B | A) = \frac{\binom{5}{3}}{\binom{6}{3}} = \frac{10}{20} = \frac{1}{2} \]
Step 3c: Probability of \( B \) (Total probability that 3 drawn balls are green) This includes two scenarios: 1. 5 green balls in the bag Probability = \( \frac{1}{3} \) and probability of drawing 3 green balls = \( \frac{10}{20} = \frac{1}{2} \) 2. 6 green balls in the bag Probability = \( \frac{1}{3} \) and probability of drawing 3 green balls = 1 By total probability law: \[ P(B) = \frac{1}{3} \times \frac{1}{2} + \frac{1}{3} \times 1 \] \[ P(B) = \frac{1}{6} + \frac{1}{3} = \frac{1 + 2}{6} = \frac{3}{6} = \frac{1}{2} \]
Step 4: Calculate \( P(A | B) \) By Bayes' theorem: \[ P(A | B) = \frac{P(B | A) \cdot P(A)}{P(B)} = \frac{\frac{1}{2} \times \frac{1}{3}}{\frac{1}{2}} \] \[ P(A | B) = \frac{\frac{1}{6}}{\frac{1}{2}} = \frac{1}{6} \times 2 = \frac{1}{3} \]
Step 5: Final Answer \[ Correct Answer: (3) \ \frac{2}{7} \]

Answer 2

To solve this problem, we use the concept of probability and combinations. We need to find the probability that exactly 5 out of the 6 balls are green, given that three drawn balls are green. Let's denote:

• \( G \) as a green ball and \( N \) as a non-green ball.

If 5 balls are green, the possible compositions of the bag are \( GGGGGN \). We consider cases where 3 balls drawn are green under these conditions.
Now, calculate probability:

1. Total ways to choose 3 balls from 6 is given by \( \binom{6}{3} \).
2. Probability of drawing 3 green balls from 5 green is \( \binom{5}{3} \).
3. Total favorable cases for this situation: \( \binom{5}{3} \times \binom{1}{0} \) choosing all 3 balls from the green ones.
4. Calculate total possible 3-ball combinations from 6 balls available to us: \[\binom{6}{3} = 20\]
5. Calculate combinations to draw 3 greens from 5 green balls: \[\binom{5}{3} = 10\]
6. Thus, probability that exactly 3 drawn balls are green if there are 5 green balls: \[\frac{\binom{5}{3}}{\binom{6}{3}} = \frac{10}{20} = \frac{1}{2}\]

Case	Probability
5 green balls	\(\frac{1}{2}\)

Given that probability of drawing 3 green balls is \(1\). Apply Bayes', considering probability as \(\text{P(5green|3 green drawn)}\):
\[\text{P(3 green drawn|5 green)} = \frac{1}{2}\]
By total probability law, \[\text{Total probability } = \frac{1}{2} \cdot \text{P(5green)} + \frac{1}{1} \cdot \text{P(6green)}=1\]
Set equal probabilities solving: \[\frac{1}{2}x + y = 1\]
Assume x and y reflect respective probabilities. Set: \[x + y = 1,\text{ and } \frac{x}{2} + y = 1 \rightarrow x = \frac{2}{3}, y = \frac{1}{3}\]
\[\text{P(5 green balls)} = \frac{2}{7}\]
The probability that exactly 5 of the balls in the bag are green, given that 3 selected ones are green, is \(\frac{2}{7}\).