Whenever horses `a , b , c` race together, their respective probabilities of winning the race are 0.3, 0.5, and 0.2 respectively. If they race three times, the pr4obability t hat the same horse wins all the three races, and the probability that `a , b ,c` each wins one race are, respectively. `8//50 ,9//50` b. `16//100 ,3//100` c. `12//50 , 15//50` d. `10//50 ,8//50`

Options

A. `8//50,9//50`
B. `16//100,3//100`
C. `12//50,15//50`
D. `10//50,8//50`

Explanation

To solve the problem step-by-step, we need to calculate two probabilities: the probability that the same horse wins all three races and the probability that each horse wins one race when they race three times.

### Step 1: Calculate the probability that the same horse wins all three races.

1. **Identify the probabilities of each horse winning a single race:**
- Probability of horse A winning = \( P(A) = 0.3 \)
- Probability of horse B winning = \( P(B) = 0.5 \)
- Probability of horse C winning = \( P(C) = 0.2 \)

2. **Calculate the probability of each horse winning all three races:**
- For horse A winning all three races:
\[ P(A \text{ wins all 3}) = P(A) \times P(A) \times P(A) = (0.3)^3 = 0.027 \]
- For horse B winning all three races:
\[ P(B \text{ wins all 3}) = P(B) \times P(B) \times P(B) = (0.5)^3 = 0.125 \]
- For horse C winning all three races:
\[ P(C \text{ wins all 3}) = P(C) \times P(C) \times P(C) = (0.2)^3 = 0.008 \]

3. **Sum the probabilities:**
\[ P(\text{same horse wins all 3}) = P(A \text{ wins all 3}) + P(B \text{ wins all 3}) + P(C \text{ wins all 3}) \]
\[ = 0.027 + 0.125 + 0.008 = 0.160 \]

4. **Convert to a fraction:**
\[ 0.160 = \frac{160}{1000} = \frac{8}{50} \]

### Step 2: Calculate the probability that each horse wins one race.

1. **Determine the number of ways each horse can win one race:**
- The arrangement of A, B, and C winning one race each can occur in \( 3! = 6 \) different ways.

2. **Calculate the probability for one specific arrangement (e.g., A wins first, B wins second, C wins third):**
\[ P(A \text{ wins 1st}) \times P(B \text{ wins 2nd}) \times P(C \text{ wins 3rd}) = 0.3 \times 0.5 \times 0.2 = 0.03 \]

3. **Multiply by the number of arrangements:**
\[ P(\text{each wins one}) = 6 \times 0.03 = 0.18 \]

4. **Convert to a fraction:**
\[ 0.18 = \frac{18}{100} = \frac{9}{50} \]

### Final Result
The probabilities are:
- Probability that the same horse wins all three races: \( \frac{8}{50} \)
- Probability that each horse wins one race: \( \frac{9}{50} \)

Thus, the answer is option **a: \( \frac{8}{50}, \frac{9}{50} \)**.