A bag contains 4 white and 6 black balls , another bag contains 4 white and 4 black balls . From any one of these bags a single draw of two balls is made .Find the probability that one of them would be white and other black ball.

Options

A. `(29)/(105)`
B. `(58)/(105)`
C. `(29)/(210)`
D. None ot these

Answer

To solve the problem step-by-step, we will calculate the probability of drawing one white ball and one black ball from each of the two bags, and then combine these probabilities.

### Step 1: Understand the contents of the bags
- **Bag A** contains 4 white balls and 6 black balls.
- **Bag B** contains 4 white balls and 4 black balls.

### Step 2: Calculate the total number of balls in each bag
- Total in Bag A = 4 (white) + 6 (black) = 10 balls.
- Total in Bag B = 4 (white) + 4 (black) = 8 balls.

### Step 3: Calculate the probability of choosing each bag
Since there are two bags, the probability of choosing either bag is:
- P(Bag A) = 1/2
- P(Bag B) = 1/2

### Step 4: Calculate the probability of drawing one white and one black ball from Bag A
To find this probability, we use combinations:
- The number of ways to choose 1 white ball from 4 white balls = \( \binom{4}{1} = 4 \).
- The number of ways to choose 1 black ball from 6 black balls = \( \binom{6}{1} = 6 \).
- The total ways to choose 2 balls from Bag A = \( \binom{10}{2} = 45 \).

Thus, the probability of drawing one white and one black ball from Bag A is:
\[ P(\text{1W, 1B | Bag A}) = \frac{\binom{4}{1} \times \binom{6}{1}}{\binom{10}{2}} = \frac{4 \times 6}{45} = \frac{24}{45} = \frac{8}{15} \]

### Step 5: Calculate the probability of drawing one white and one black ball from Bag B
Similarly, for Bag B:
- The number of ways to choose 1 white ball from 4 white balls = \( \binom{4}{1} = 4 \).
- The number of ways to choose 1 black ball from 4 black balls = \( \binom{4}{1} = 4 \).
- The total ways to choose 2 balls from Bag B = \( \binom{8}{2} = 28 \).

Thus, the probability of drawing one white and one black ball from Bag B is:
\[ P(\text{1W, 1B | Bag B}) = \frac{\binom{4}{1} \times \binom{4}{1}}{\binom{8}{2}} = \frac{4 \times 4}{28} = \frac{16}{28} = \frac{4}{7} \]

### Step 6: Combine the probabilities
Now, we combine the probabilities from both bags using the law of total probability:
\[ P(\text{1W, 1B}) = P(\text{Bag A}) \times P(\text{1W, 1B | Bag A}) + P(\text{Bag B}) \times P(\text{1W, 1B | Bag B}) \]
\[ P(\text{1W, 1B}) = \left(\frac{1}{2} \times \frac{8}{15}\right) + \left(\frac{1}{2} \times \frac{4}{7}\right) \]
\[ = \frac{8}{30} + \frac{4}{14} \]
To add these fractions, we need a common denominator. The least common multiple of 30 and 14 is 210.
\[ = \frac{8 \times 7}{210} + \frac{4 \times 15}{210} = \frac{56}{210} + \frac{60}{210} = \frac{116}{210} \]
Simplifying this gives:
\[ = \frac{58}{105} \]

### Final Answer
The probability that one of the balls drawn is white and the other is black is \( \frac{58}{105} \).

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