There are 5 red and 7 white balls in one bag and 3 red and 8 white balls in second bg. If a ball is drawn at random from one the bags, find the probability that it is red.

Explanation

To solve the problem of finding the probability of drawing a red ball from one of the two bags, we can follow these steps:

### Step 1: Determine the total number of balls in each bag.
- **Bag 1**: 5 red balls + 7 white balls = 12 balls
- **Bag 2**: 3 red balls + 8 white balls = 11 balls

### Step 2: Calculate the probability of drawing a red ball from Bag 1.
- The probability \( P(\text{Red from Bag 1}) \) is given by the formula:
\[ P(\text{Red from Bag 1}) = \frac{\text{Number of Red Balls in Bag 1}}{\text{Total Number of Balls in Bag 1}} = \frac{5}{12} \]

### Step 3: Calculate the probability of drawing a red ball from Bag 2.
- The probability \( P(\text{Red from Bag 2}) \) is given by the formula:
\[ P(\text{Red from Bag 2}) = \frac{\text{Number of Red Balls in Bag 2}}{\text{Total Number of Balls in Bag 2}} = \frac{3}{11} \]

### Step 4: Calculate the total probability of drawing a red ball.
- Since we can draw from either bag, we need to find the combined probability of drawing a red ball from either bag. Assuming that each bag is equally likely to be chosen, we can use the law of total probability:
\[ P(\text{Red}) = P(\text{Red from Bag 1}) \cdot P(\text{Bag 1}) + P(\text{Red from Bag 2}) \cdot P(\text{Bag 2}) \]
- Assuming both bags are equally likely to be chosen, \( P(\text{Bag 1}) = P(\text{Bag 2}) = \frac{1}{2} \):
\[ P(\text{Red}) = \left(\frac{5}{12} \cdot \frac{1}{2}\right) + \left(\frac{3}{11} \cdot \frac{1}{2}\right) \]
\[ P(\text{Red}) = \frac{5}{24} + \frac{3}{22} \]

### Step 5: Find a common denominator and add the probabilities.
- The least common multiple of 24 and 22 is 264.
- Convert each fraction:
\[ \frac{5}{24} = \frac{5 \times 11}{24 \times 11} = \frac{55}{264} \]
\[ \frac{3}{22} = \frac{3 \times 12}{22 \times 12} = \frac{36}{264} \]
- Now add the two fractions:
\[ P(\text{Red}) = \frac{55}{264} + \frac{36}{264} = \frac{91}{264} \]

### Final Answer:
The probability of drawing a red ball is \( \frac{91}{264} \).

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