Box A has 3 white and 2 red balls, box B has 2 white and 4 red balls. If two balls are selected at random without replacements from the box A and 2 more are selected at random from B, the probability that all the four balls are white is

Options

A. `1/10`
B. `1/50`
C. `1-((.^(2)C_(2))/(.^(5)C_(2)).(.^(4)C_(2))/(.^(5)C_(2)))`
D. `(.^(3)C_(2).^(2)C_(2))/(.^(5)C_(2).^(6)C_(2))`

Answer

To solve the problem step by step, we need to find the probability of selecting 4 white balls when 2 are drawn from Box A and 2 from Box B.

### Step 1: Understand the contents of the boxes
- Box A contains 3 white balls and 2 red balls.
- Box B contains 2 white balls and 4 red balls.

### Step 2: Calculate the probability of drawing 2 white balls from Box A
The total number of ways to choose 2 balls from Box A (which has 5 balls in total) is given by the combination formula \( \binom{n}{r} \), where \( n \) is the total number of balls and \( r \) is the number of balls to choose.

The number of ways to choose 2 white balls from the 3 white balls in Box A is \( \binom{3}{2} \).

The total combinations of choosing any 2 balls from Box A is \( \binom{5}{2} \).

So, the probability \( P(A) \) of drawing 2 white balls from Box A is:

\[ P(A) = \frac{\binom{3}{2}}{\binom{5}{2}} = \frac{3}{10} \]

### Step 3: Calculate the probability of drawing 2 white balls from Box B
The number of ways to choose 2 white balls from the 2 white balls in Box B is \( \binom{2}{2} \).

The total combinations of choosing any 2 balls from Box B (which has 6 balls in total) is \( \binom{6}{2} \).

So, the probability \( P(B) \) of drawing 2 white balls from Box B is:

\[ P(B) = \frac{\binom{2}{2}}{\binom{6}{2}} = \frac{1}{15} \]

### Step 4: Calculate the combined probability of both events
Since the selections from Box A and Box B are independent events, the combined probability \( P(A \text{ and } B) \) of both events happening (drawing 2 white balls from Box A and 2 white balls from Box B) is given by:

\[ P(A \text{ and } B) = P(A) \times P(B) = \frac{3}{10} \times \frac{1}{15} \]

Calculating this gives:

\[ P(A \text{ and } B) = \frac{3}{150} = \frac{1}{50} \]

### Final Answer
Thus, the probability that all four balls drawn are white is:

\[ \boxed{\frac{1}{50}} \]