There are 12 white and 12 red ball in a bag. Balls are drawn one by one with replacement from the bag. The probability that 7th drawn ball is 4th white, is

Options

A. `1//4`
B. `5//32`
C. `3//16`
D. `5//16`

Explanation

To find the probability that the 7th drawn ball is the 4th white ball, we can follow these steps:

### Step 1: Understand the problem
We have a total of 24 balls (12 white and 12 red). We are drawing balls with replacement, and we want the 7th ball drawn to be the 4th white ball.

### Step 2: Determine the probabilities
The probability of drawing a white ball (P) is:
\[ P(\text{White}) = \frac{12}{24} = \frac{1}{2} \]
Similarly, the probability of drawing a red ball (Q) is:
\[ P(\text{Red}) = \frac{12}{24} = \frac{1}{2} \]

### Step 3: Set up the conditions
For the 7th ball to be the 4th white ball, we need exactly 3 white balls to be drawn in the first 6 draws.

### Step 4: Use the binomial probability formula
The number of ways to choose 3 white balls from the first 6 draws can be calculated using the binomial coefficient:
\[ \binom{6}{3} \]
The probability of drawing 3 white balls and 3 red balls in the first 6 draws is given by:
\[ \binom{6}{3} \cdot P^3 \cdot Q^3 \]
Substituting the values of P and Q:
\[ = \binom{6}{3} \cdot \left(\frac{1}{2}\right)^3 \cdot \left(\frac{1}{2}\right)^3 \]
\[ = \binom{6}{3} \cdot \left(\frac{1}{2}\right)^6 \]

### Step 5: Calculate the binomial coefficient
The binomial coefficient \(\binom{6}{3}\) is calculated as:
\[ \binom{6}{3} = \frac{6!}{3!(6-3)!} = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} = 20 \]

### Step 6: Substitute and simplify
Now substitute back into the probability expression:
\[ = 20 \cdot \left(\frac{1}{2}\right)^6 = 20 \cdot \frac{1}{64} = \frac{20}{64} = \frac{5}{16} \]

### Step 7: Include the probability of the 7th ball being white
Since the 7th ball must also be white, we multiply the above result by the probability of drawing a white ball:
\[ = \frac{5}{16} \cdot \frac{1}{2} = \frac{5}{32} \]

### Final Answer
Thus, the probability that the 7th drawn ball is the 4th white ball is:
\[ \frac{5}{32} \]

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