In a diving competition, each diver has a `20%` change of a perfect dive. The first perfect dive of the competition, but no subsequent dives, will receive a perfect score. IF Janet is the third diver to dive, what is her chance of receiving a perfect score?
(Assume that each diver can perform only one dive per turn)

Options

A. 0.5
B. 0.625
C. 0.125
D. 0.25

Answer

To solve the problem, we need to calculate the probability that Janet, the third diver, receives a perfect score. For Janet to receive a perfect score, the following conditions must be met:

1. The first diver must not achieve a perfect dive (imperfect dive).
2. The second diver must also not achieve a perfect dive (imperfect dive).
3. Janet, the third diver, must achieve a perfect dive.

Let's denote:
- The probability of a perfect dive = \( P(\text{Perfect}) = 0.2 \)
- The probability of an imperfect dive = \( P(\text{Imperfect}) = 1 - P(\text{Perfect}) = 1 - 0.2 = 0.8 \)

Now, we can calculate the probability step by step:

### Step 1: Calculate the probability of the first diver not achieving a perfect dive.
The probability that the first diver does not achieve a perfect dive is:
\[ P(\text{Imperfect for Diver 1}) = 0.8 \]

### Step 2: Calculate the probability of the second diver not achieving a perfect dive.
The probability that the second diver does not achieve a perfect dive is:
\[ P(\text{Imperfect for Diver 2}) = 0.8 \]

### Step 3: Calculate the probability of Janet achieving a perfect dive.
The probability that Janet achieves a perfect dive is:
\[ P(\text{Perfect for Janet}) = 0.2 \]

### Step 4: Combine the probabilities.
To find the total probability that Janet receives a perfect score, we multiply the probabilities from the three steps:
\[ P(\text{Janet receives perfect score}) = P(\text{Imperfect for Diver 1}) \times P(\text{Imperfect for Diver 2}) \times P(\text{Perfect for Janet}) \]
Substituting the values:
\[ P(\text{Janet receives perfect score}) = 0.8 \times 0.8 \times 0.2 \]

### Step 5: Calculate the final probability.
\[ P(\text{Janet receives perfect score}) = 0.8 \times 0.8 = 0.64 \]
\[ P(\text{Janet receives perfect score}) = 0.64 \times 0.2 = 0.128 \]

Thus, the probability that Janet receives a perfect score is \( \boxed{0.128} \).