A bag initially contains 1 red and 2 blue balls. An experiment consisting of selecting a ball at random, noting its color and replacing it together with an additional ball of the same colour. If three such trials are made, then

Options

A. probability that at least one blue balls is drawn is `0.9`
B. probability that exactly one blue all is drawn is `0.2`
C. probability that all the drawn balls are red given that all the drawn balls are of same color is `0.2`
D. probability that at least one red ball is drawn is `0.6`

Explanation

To solve the problem step by step, we will analyze the situation and calculate the required probabilities based on the given conditions.

### Step 1: Understanding the Initial Setup
The bag initially contains:
- 1 Red Ball (R)
- 2 Blue Balls (B)

### Step 2: Probability of Drawing Balls
When a ball is drawn, it is noted and replaced along with an additional ball of the same color. This affects the probabilities in subsequent trials.

### Step 3: Total Balls After Each Trial
1. **Trial 1**:
- Possible outcomes:
- Draw Red (R): Bag now has 2R, 2B
- Draw Blue (B): Bag now has 1R, 3B

2. **Trial 2**:
- If R was drawn in Trial 1 (2R, 2B):
- Draw R: Bag now has 3R, 2B
- Draw B: Bag now has 2R, 3B
- If B was drawn in Trial 1 (1R, 3B):
- Draw R: Bag now has 2R, 3B
- Draw B: Bag now has 1R, 4B

3. **Trial 3**: Similar logic applies, but we will calculate probabilities based on the outcomes of the previous trials.

### Step 4: Probability of Drawing at Least One Blue Ball
To find the probability of drawing at least one blue ball in three trials, we can use the complement rule:
\[ P(\text{At least one blue}) = 1 - P(\text{No blue}) \]

- The only way to have no blue balls is to draw red balls in all three trials.
- The probability of drawing red balls in each trial will change based on previous outcomes.

1. **Probability of drawing RRR**:
- For the first draw: P(R) = 1/3
- For the second draw (if R was drawn first): P(R) = 2/4 = 1/2
- For the third draw (if RR was drawn): P(R) = 3/5

Thus,
\[ P(RRR) = \frac{1}{3} \times \frac{1}{2} \times \frac{3}{5} = \frac{1}{10} \]

Now, calculate the probability of at least one blue:
\[ P(\text{At least one blue}) = 1 - P(RRR) = 1 - \frac{1}{10} = \frac{9}{10} = 0.9 \]

### Step 5: Probability of Exactly One Blue Ball
To find the probability of exactly one blue ball being drawn, we can consider the combinations of drawing one blue and two reds in any order.

The favorable outcomes are:
1. R, R, B
2. R, B, R
3. B, R, R

Calculating each:
- For RRB:
\[ P(RRB) = \frac{1}{3} \times \frac{1}{2} \times \frac{3}{5} = \frac{1}{10} \]
- For RBR:
\[ P(RBR) = \frac{1}{3} \times \frac{2}{4} \times \frac{3}{5} = \frac{1}{10} \]
- For BRR:
\[ P(BRR) = \frac{2}{3} \times \frac{2}{4} \times \frac{3}{5} = \frac{1}{10} \]

Total probability for exactly one blue:
\[ P(\text{Exactly one blue}) = 3 \times \frac{1}{10} = \frac{3}{10} = 0.2 \]

### Step 6: Probability that All Drawn Balls are Red Given All are the Same Color
Here we need to find:
\[ P(RRR | \text{All same color}) \]

The possible outcomes for all same color are RRR and BBB.
- We already calculated P(RRR) = 1/10.
- Now we need P(BBB):
1. For the first draw: P(B) = 2/3
2. For the second draw (if B was drawn): P(B) = 3/4
3. For the third draw (if BB was drawn): P(B) = 4/5

Thus,
\[ P(BBB) = \frac{2}{3} \times \frac{3}{4} \times \frac{4}{5} = \frac{24}{60} = \frac{2}{5} \]

Now, total probability for all same color:
\[ P(\text{All same color}) = P(RRR) + P(BBB) = \frac{1}{10} + \frac{2}{5} = \frac{1}{10} + \frac{4}{10} = \frac{5}{10} = \frac{1}{2} \]

Finally,
\[ P(RRR | \text{All same color}) = \frac{P(RRR)}{P(\text{All same color})} = \frac{\frac{1}{10}}{\frac{1}{2}} = \frac{1}{5} = 0.2 \]

### Step 7: Probability of At Least One Red Ball
Using the complement rule again:
\[ P(\text{At least one red}) = 1 - P(BBB) \]
We already calculated P(BBB) = 2/5.
Thus,
\[ P(\text{At least one red}) = 1 - \frac{2}{5} = \frac{3}{5} = 0.6 \]

### Summary of Results
1. Probability that at least one blue is drawn: **0.9**
2. Probability that exactly one blue ball is drawn: **0.2**
3. Probability that all drawn balls are red given that all drawn balls are of the same color: **0.2**
4. Probability that at least one red ball is drawn: **0.6**