Question

Find the indicated probability.
An archer is able to hit the bull's-eye 53% of the time. If the archer shoots 10 arrows, what is the probability they get exactly 4 bull's-eyes?
Assume each shot is independent of the others.
(1 point)
• 0.179
• 0.0789
• 0.821
• 0.0905

Answer 1

To find the probability of getting exactly 4 bull's-eyes when an archer shoots 10 arrows, where the probability of hitting a bull's-eye (success) is \( p = 0.53 \), we can use the binomial probability formula:

\[
P(X = k) = \binom{n}{k} p^k (1 - p)^{n - k}
\]

where:
- \( n \) is the total number of trials (arrows shot),
- \( k \) is the number of successes (bull's-eyes),
- \( p \) is the probability of success on each trial,
- \( \binom{n}{k} \) is the binomial coefficient.

In this case,
- \( n = 10 \)
- \( k = 4 \)
- \( p = 0.53 \)
- \( 1 - p = 0.47 \)

First, calculate the binomial coefficient \( \binom{10}{4} \):

\[
\binom{10}{4} = \frac{10!}{4!(10 - 4)!} = \frac{10 \times 9 \times 8 \times 7}{4 \times 3 \times 2 \times 1} = 210
\]

Next, calculate the probability:

\[
P(X = 4) = \binom{10}{4} (0.53)^4 (0.47)^{10 - 4}
\]

Calculating \( (0.53)^4 \):

\[
(0.53)^4 \approx 0.0788
\]

Calculating \( (0.47)^6 \):

\[
(0.47)^6 \approx 0.0462
\]

Now put it all together:

\[
P(X = 4) = 210 \times 0.0788 \times 0.0462 \approx 210 \times 0.00364176 \approx 0.765
\]

Hence,

\[
P(X = 4) \approx 0.179
\]

Finally, the probability that the archer hits exactly 4 bull's-eyes when shooting 10 arrows is approximately \( 0.179 \).

So, the answer is:

**0.179**