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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.
0.0789
0.0905
0.179
0.821

Sagot :

Answer:

C) 0.179

Step-by-step explanation:

Since the trials are independent, this is a binomial distribution:

Recall:

  • Binomial Distribution --> [tex]P(k)={n\choose k}p^kq^{n-k}[/tex]
  • [tex]P(k)[/tex] denotes the probability of [tex]k[/tex] successes in [tex]n[/tex] independent trials
  • [tex]p^k[/tex] denotes the probability of success on each of [tex]k[/tex] trials
  • [tex]q^{n-k}[/tex] denotes the probability of failure on the remaining [tex]n-k[/tex] trials
  • [tex]{n\choose k}=\frac{n!}{(n-k)!k!}[/tex] denotes all possible ways to choose [tex]k[/tex] things out of [tex]n[/tex] things

Given:

  • [tex]n=10[/tex]
  • [tex]k=4[/tex]
  • [tex]p^k=0.53^4[/tex]
  • [tex]q^{n-k}=(1-0.53)^{10-4}=0.47^6[/tex]
  • [tex]{n\choose k}={10\choose 4}=\frac{10!}{(10-4)!4!}=210[/tex]

Calculate:

  • [tex]P(4)=(210)(0.53^4)(0.47^6)=0.1786117069\approx0.179[/tex]

Therefore, the probability that the archer will get exactly 4 bull's-eyes with 10 arrows in any order is 0.179

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