Bag A contains three red and four white Balls , bag B contains two red and three white Balls. If one Ball is drawn from bag A and two Balls from bag B find the probability that : One Ball is red and two Balls are white .

Answer

To solve the problem step by step, we will calculate the probability of drawing one red ball from Bag A and two white balls from Bag B.

### Step 1: Determine the total number of balls in each bag.
- **Bag A** contains 3 red and 4 white balls.
- Total balls in Bag A = 3 + 4 = 7
- **Bag B** contains 2 red and 3 white balls.
- Total balls in Bag B = 2 + 3 = 5

### Step 2: Calculate the probability of drawing one red ball from Bag A.
- The number of ways to choose 1 red ball from 3 red balls is given by \( \binom{3}{1} \).
- The total number of ways to choose 1 ball from 7 balls in Bag A is \( \binom{7}{1} \).

\[ P(\text{Red from A}) = \frac{\binom{3}{1}}{\binom{7}{1}} = \frac{3}{7} \]

### Step 3: Calculate the probability of drawing two white balls from Bag B.
- The number of ways to choose 2 white balls from 3 white balls is given by \( \binom{3}{2} \).
- The total number of ways to choose 2 balls from 5 balls in Bag B is \( \binom{5}{2} \).

\[ P(\text{2 White from B}) = \frac{\binom{3}{2}}{\binom{5}{2}} = \frac{3}{10} \]

### Step 4: Calculate the combined probability of the events.
Since the events of drawing from Bag A and Bag B are independent, we can multiply the probabilities:

\[ P(\text{1 Red from A and 2 White from B}) = P(\text{Red from A}) \times P(\text{2 White from B}) \]

Substituting the values we calculated:

\[ P(\text{1 Red from A and 2 White from B}) = \frac{3}{7} \times \frac{3}{10} = \frac{9}{70} \]

### Step 5: Calculate the probability of drawing one white ball from Bag A and one red ball from Bag B.
- The number of ways to choose 1 white ball from 4 white balls in Bag A is \( \binom{4}{1} \).
- The number of ways to choose 1 red ball from 2 red balls in Bag B is \( \binom{2}{1} \).

\[ P(\text{1 White from A}) = \frac{\binom{4}{1}}{\binom{7}{1}} = \frac{4}{7} \]
\[ P(\text{1 Red from B}) = \frac{\binom{2}{1}}{\binom{5}{1}} = \frac{2}{5} \]

### Step 6: Calculate the combined probability of this second scenario.
\[ P(\text{1 White from A and 1 Red from B}) = P(\text{1 White from A}) \times P(\text{1 Red from B}) = \frac{4}{7} \times \frac{2}{5} = \frac{8}{35} \]

### Final Step: Combine both probabilities.
The total probability of getting one red ball and two white balls (either from Bag A or Bag B) is:

\[ P(\text{Total}) = P(\text{1 Red from A and 2 White from B}) + P(\text{1 White from A and 1 Red from B}) = \frac{9}{70} + \frac{8}{35} \]

To add these fractions, we need a common denominator:

\[ \frac{8}{35} = \frac{16}{70} \]

Thus,

\[ P(\text{Total}) = \frac{9}{70} + \frac{16}{70} = \frac{25}{70} = \frac{5}{14} \]

### Conclusion
The probability that one ball is red and two balls are white is \( \frac{5}{14} \).

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