2 balls are selected with replacement from 10 red, 30 white, 15 orange and 20 blue balls then probability that first ball is red and second ball is white, is.

Options

A. `9/25`
B. `4/75`
C. `8/75`
D. `7/75`

Answer

To solve the problem of finding the probability that the first ball selected is red and the second ball selected is white, we can follow these steps:

### Step 1: Determine the total number of balls
We have the following counts of balls:
- Red balls = 10
- White balls = 30
- Orange balls = 15
- Blue balls = 20

To find the total number of balls, we add these together:
\[ \text{Total balls} = 10 + 30 + 15 + 20 = 75 \]

### Step 2: Calculate the probability of selecting a red ball first
The probability of selecting a red ball from the total is given by the ratio of red balls to the total number of balls:
\[ P(\text{Red}) = \frac{\text{Number of Red Balls}}{\text{Total Number of Balls}} = \frac{10}{75} \]

### Step 3: Calculate the probability of selecting a white ball second
Since we are selecting with replacement, the total number of balls remains the same for the second selection. The probability of selecting a white ball is:
\[ P(\text{White}) = \frac{\text{Number of White Balls}}{\text{Total Number of Balls}} = \frac{30}{75} \]

### Step 4: Calculate the combined probability
Since the selections are independent events (due to replacement), we can multiply the probabilities of the two events:
\[ P(\text{Red first and White second}) = P(\text{Red}) \times P(\text{White}) = \left(\frac{10}{75}\right) \times \left(\frac{30}{75}\right) \]

### Step 5: Simplify the expression
Calculating the product:
\[ P(\text{Red first and White second}) = \frac{10 \times 30}{75 \times 75} = \frac{300}{5625} \]

Now, simplifying \(\frac{300}{5625}\):
\[ \frac{300 \div 75}{5625 \div 75} = \frac{4}{75} \]

### Final Answer
Thus, the probability that the first ball is red and the second ball is white is:
\[ \boxed{\frac{4}{75}} \]

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