१६. एउटा बाकसमा 6 ओटा राता र ४ ओटा सेता उस्तै र उत्रै बलहरू राखिएका छन् ।

उक्त बाकसबाट दुइओटा बलहरू एकपछि अर्को नहेरीकन (पुनः नराखिकन) निकालिएका छन् ।

A box contains 6 red and 8 white balls of the same shape and size. Two balls are drawn randomly one after another without replacement

from the box.

क) A र B दुईओटा अनाश्रित घटनाहरू हुन् भने P(A∩B) पत्ता लगाउने सुत्र ( लेख्नुहोस् ।

(१) A and B are two independent events, write the formula to find P(A∩B).

(ख) माथिको कथनबाट प्राप्त हुने सम्भावित परिणामहरूको सम्भाव्यतालाई वृक्ष चित्रमा देखाउनुहोस् । (२)

Show the probability of possible outcomes of the above statement in a tree diagram.

(ग) दुवै वल एउटै रङका पर्ने सम्भाव्यता पत्ता लगाउनुहोस् । Find the probability of getting both balls of same colour. (१)

घ) दुवै वल एउटै रङ्का पर्ने सम्भाव्यता र दुवै वल फरक रङ्का पर्ने सम्भाव्यता

मध्ये कुन चाहिँको सम्भाव्यता बढी हुन्छ ? गणना गर्नुहोस् । (१) Which one probability is more either both ball are same colour or both balls are different colour? Calculate it.

Let's go step by step to solve the problem:

### Given:
- 6 red balls
- 4 white balls
- Two balls are drawn one after another without replacement.

### (क) **Formula to find P(A∩B) if A and B are independent events**:
The formula to find the probability of the intersection of two independent events \(A\) and \(B\) is given by:
\[
P(A \cap B) = P(A) \times P(B)
\]

### (ख) **Tree Diagram**:
To illustrate the probability of possible outcomes, we can draw a tree diagram. Each branch represents a possible outcome, and its associated probability is shown.

1. **First draw (First Ball)**:
    - Red Ball: Probability = \(\frac{6}{10} = 0.6\)
    - White Ball: Probability = \(\frac{4}{10} = 0.4\)

2. **Second draw (Second Ball, based on first draw)**:
   - **If the first ball is red**:
     - Red Ball: Probability = \(\frac{5}{9} \approx 0.555\)
     - White Ball: Probability = \(\frac{4}{9} \approx 0.444\)
   - **If the first ball is white**:
     - Red Ball: Probability = \(\frac{6}{9} = \frac{2}{3} \approx 0.666\)
     - White Ball: Probability = \(\frac{3}{9} = \frac{1}{3} \approx 0.333\)

### (ग) **Find the probability of getting both balls of the same color**:
- **Case 1: Both balls are red**:
  \[
  P(\text{Both Red}) = \frac{6}{10} \times \frac{5}{9} = \frac{30}{90} = \frac{1}{3} \approx 0.333
  \]

- **Case 2: Both balls are white**:
  \[
  P(\text{Both White}) = \frac{4}{10} \times \frac{3}{9} = \frac{12}{90} = \frac{2}{15} \approx 0.133
  \]

- **Total probability of both balls being of the same color**:
  \[
  P(\text{Same Color}) = \frac{1}{3} + \frac{2}{15} = \frac{5}{15} \approx 0.467
  \]

### (घ) **Which probability is more: Both balls are of the same color or both balls are of different colors?**:
- **Probability of different colors**:
  \[
  P(\text{Different Colors}) = 1 - P(\text{Same Color}) = 1 - 0.467 = 0.533
  \]

Since \(P(\text{Different Colors}) = 0.533\) and \(P(\text{Same Color}) = 0.467\), the probability of drawing two balls of different colors is higher.

---

Do you want details or have any questions? 

**Related Questions:**
1. What is the probability of drawing two red balls in the first two draws?
2. If one ball is replaced after the first draw, how does the probability tree diagram change?
3. What would be the probability of getting one red and one white ball if the order didn't matter?
4. How would the probabilities change if the number of balls were different?
5. Can you extend this problem to three draws without replacement?
6. What is the expected number of red balls drawn after two draws?
7. How does conditional probability relate to this problem?
8. What happens if the balls are not identical in shape and size?

**Tip:** Understanding the concept of independent and dependent events is crucial for solving problems involving probability trees.

Understand the probability of drawing two balls from a box containing 6 red and 4 white balls, without replacement. Learn the formulas for calculating probabilities of independent events and explore tree diagrams for visualizing outcomes.

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