A kicker for a football team makes field goals [tex]$85\%$[/tex] of the time when kicking from the 20-yard line. Assume that field-goal attempts can be considered random events. Using the table, what is the probability that the placekicker will make at least 4 of his next 5 attempts from the 20-yard line?

\begin{tabular}{|l|c|c|c|c|c|c|}
\hline Field Goals Made & 0 & 1 & 2 & 3 & 4 & 5 \\
\hline Probability & 0.00008 & 0.002 & 0.024 & 0.138 & 0.392 & ? \\
\hline
\end{tabular}

A. 0.392
B. 0.444
C. 0.556
D. 0.836



Answer :

To find the probability that the placekicker will make at least 4 of his next 5 attempts from the 20-yard line, we'll consider the probabilities for making exactly 4 and exactly 5 field goals.

First, let's look at the probabilities given in the table:
- The probability of making exactly 4 field goals is 0.392.
- We need to determine the probability of making exactly 5 field goals.

### Step-by-Step Explanation:

1. Extract the Given Information:
- The probability of making exactly 4 field goals out of 5 is [tex]\( P(4) = 0.392 \)[/tex].
- The total number of attempts is [tex]\( n = 5 \)[/tex].
- The probabilities for making fewer than 4 field goals are provided but are not directly necessary for calculating the desired outcome.

2. Probability of Making Exactly 5 Field Goals:
- The given problem requires the probability of making at least 4 out of 5 attempts.
- [tex]\( P(5) = 0.444 \)[/tex] is the probability of making 5 field goals.

3. Summing the Probabilities:
- The probability of making at least 4 out of 5 is the sum of the probabilities of making exactly 4 and exactly 5 field goals.
- Thus, we need to add [tex]\( P(4) \)[/tex] and [tex]\( P(5) \)[/tex].

[tex]\[ P(\text{at least 4}) = P(4) + P(5) = 0.392 + 0.444 = 0.836 \][/tex]

4. Conclusion:
- Therefore, the probability that the placekicker will make at least 4 of his next 5 attempts from the 20-yard line is:

[tex]\[ 0.836 \][/tex]

This matches one of the choices provided. The correct answer is indeed:

[tex]\[ 0.836 \][/tex]