The question involves calculating the probability that a soccer player scores a goal once and misses once, in two independent attempts, given the probability of scoring in one attempt is $0.7$.

Solution:

1. Let the probability of scoring in one attempt be $P(\text{score}) = 0.7$. Then, the probability of missing is $P(\text{miss}) = 1 - 0.7 = 0.3$.

2. The player needs to score once and miss once. This can happen in two ways:

   • The player scores first, then misses.
   • The player misses first, then scores.

3. Using the rule for independent events, the probability of each sequence is: $P(\text{score then miss}) = P(\text{score}) \times P(\text{miss}) = 0.7 \times 0.3 = 0.21$ $P(\text{miss then score}) = P(\text{miss}) \times P(\text{score}) = 0.3 \times 0.7 = 0.21$

4. Add the probabilities for the two cases: $P(\text{one score and one miss}) = 0.21 + 0.21 = 0.42$

Answer:

The probability that the player scores one goal and misses one goal is $\mathbf{0.42}$. The correct option is B.

Would you like a more detailed explanation or further clarification?

Here are 5 related questions to expand your understanding:

1. How would the result change if the probability of scoring was $0.5$?
2. What is the probability of scoring both attempts instead of just one?
3. How can this problem be generalized for $n$ independent attempts?
4. What is the total probability of missing at least one goal out of two attempts?
5. How does independence of events influence the calculation?

Tip: When events are independent, their joint probability is the product of their individual probabilities.