A bag contains 5 white, 4 black and 2 red balls. Balls are drawn one by one without replacement. The probability that the `5^("th")` ball is a red ball, is

Options

A. `(2)/(11)`
B. `(4)/(11)`
C. `(3)/(7)`
D. `(6)/(11)`

Answer

To find the probability that the 5th ball drawn from a bag containing 5 white, 4 black, and 2 red balls is a red ball, we can follow these steps:

### Step 1: Understand the scenario
We have a total of 11 balls:
- 5 white
- 4 black
- 2 red

We want to find the probability that the 5th ball drawn is red.

### Step 2: Determine the conditions
For the 5th ball to be red, the first four balls must be drawn from the remaining balls (i.e., not red). This means that the first four balls can only be white or black.

### Step 3: Calculate the total ways to choose the first four balls
The total number of ways to choose any 4 balls from the 11 balls is given by:
\[ \text{Total ways} = \binom{11}{4} \]

### Step 4: Calculate the ways to choose 4 balls that are not red
The number of non-red balls is 9 (5 white + 4 black). The number of ways to choose 4 balls from these 9 is:
\[ \text{Ways to choose 4 non-red balls} = \binom{9}{4} \]

### Step 5: Calculate the probability that the 5th ball is red
Now, if the first four balls are chosen from the 9 non-red balls, we still have 2 red balls left. The probability that the 5th ball drawn is red can be calculated as follows:
\[ P(\text{5th ball is red}) = \frac{\text{Ways to choose 4 non-red balls} \times \text{Ways to choose 1 red ball}}{\text{Total ways to choose 4 balls}} \]

Since there are 2 red balls left, the number of ways to choose 1 red ball is:
\[ \text{Ways to choose 1 red ball} = 2 \]

Putting it all together, we have:
\[ P(\text{5th ball is red}) = \frac{\binom{9}{4} \times 2}{\binom{11}{4}} \]

### Step 6: Calculate the values of the combinations
Now we need to calculate the values of the combinations:
\[ \binom{11}{4} = \frac{11 \times 10 \times 9 \times 8}{4 \times 3 \times 2 \times 1} = 330 \]
\[ \binom{9}{4} = \frac{9 \times 8 \times 7 \times 6}{4 \times 3 \times 2 \times 1} = 126 \]

### Step 7: Substitute the values into the probability formula
Now substituting these values into our probability formula:
\[ P(\text{5th ball is red}) = \frac{126 \times 2}{330} = \frac{252}{330} \]

### Step 8: Simplify the fraction
Now we simplify the fraction:
\[ P(\text{5th ball is red}) = \frac{252 \div 6}{330 \div 6} = \frac{42}{55} \]

### Final Answer
Thus, the probability that the 5th ball drawn is a red ball is:
\[ \frac{42}{55} \]