To calculate the probability that the kicker makes both his first and second field goals, we can approach it step by step.

Step 1: Probability of making the first kick.

The probability that the kicker makes his first kick is given directly as 64%, or 0.64.

Step 2: Probability of making the second kick given the first was made.

If he makes his first attempt, his success rate increases to 92%. Therefore, the probability of making the second kick, given that he made the first, is 92%, or 0.92.

Step 3: Total probability of making both kicks.

The total probability is the product of the two probabilities: $P(\text{first and second}) = P(\text{first}) \times P(\text{second | first})$ $P(\text{first and second}) = 0.64 \times 0.92$ $P(\text{first and second}) = 0.5888$

Final Answer:

The probability that the kicker makes both his first and second kicks is 0.589 (rounded to three decimal places).

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Would you like further clarification or details on the steps? Here are some related questions for practice:

1. What is the probability of the kicker missing his first field goal attempt?
2. If the kicker makes the first two kicks, what is the probability he will miss the third?
3. How would the probability change if the kicker's success rate on the second attempt were 80% instead of 92%?
4. What is the probability of the kicker making exactly one of the first two kicks?
5. How do you calculate the probability of multiple independent events happening in sequence?

Tip: When calculating the probability of successive events (like making two kicks in a row), always remember to multiply the probabilities of each event, especially when the events are conditional (i.e., the outcome of one depends on the other).