There are 5 red, 6 black and 5 blue balls in a bag. Out of these balls, four balls are picked at random from the bag. Then, what is the probability that one ball is red, two are black and one is blue?

Options

A. `75/362`
B. `75/364`
C. `71/362`
D. `70/363`

Answer

To solve the problem of finding the probability of picking 1 red ball, 2 black balls, and 1 blue ball from a bag containing 5 red, 6 black, and 5 blue balls, we can follow these steps:

### Step 1: Determine the total number of balls
We have:
- Red balls = 5
- Black balls = 6
- Blue balls = 5

Total number of balls = 5 + 6 + 5 = 16

### Step 2: Calculate the number of ways to select the required balls
We need to select:
- 1 red ball
- 2 black balls
- 1 blue ball

**Number of ways to select 1 red ball:**
\[ \text{Number of ways} = \binom{5}{1} = 5 \]

**Number of ways to select 2 black balls:**
\[ \text{Number of ways} = \binom{6}{2} = \frac{6!}{2!(6-2)!} = \frac{6 \times 5}{2 \times 1} = 15 \]

**Number of ways to select 1 blue ball:**
\[ \text{Number of ways} = \binom{5}{1} = 5 \]

### Step 3: Calculate the total number of favorable outcomes
Now, we multiply the number of ways to select each type of ball:
\[ \text{Total favorable outcomes} = \binom{5}{1} \times \binom{6}{2} \times \binom{5}{1} = 5 \times 15 \times 5 = 375 \]

### Step 4: Calculate the total number of ways to select any 4 balls from 16
The total number of ways to select 4 balls from 16 is given by:
\[ \text{Total outcomes} = \binom{16}{4} = \frac{16!}{4!(16-4)!} = \frac{16 \times 15 \times 14 \times 13}{4 \times 3 \times 2 \times 1} = 1820 \]

### Step 5: Calculate the probability
The probability \( P \) of selecting 1 red ball, 2 black balls, and 1 blue ball is given by the ratio of favorable outcomes to total outcomes:
\[ P = \frac{\text{Total favorable outcomes}}{\text{Total outcomes}} = \frac{375}{1820} \]

### Step 6: Simplify the probability
To simplify \( \frac{375}{1820} \):
- The greatest common divisor (GCD) of 375 and 1820 is 25.
- Dividing both the numerator and the denominator by 25 gives:
\[ P = \frac{15}{72.8} \approx \frac{75}{364} \]

### Final Answer
Thus, the probability that one ball is red, two are black, and one is blue is:
\[ \frac{75}{364} \]