Question

Find the indicated probability.

Answer 1

Answer:

C) 0.179

Step-by-step explanation:

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

Recall:

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

Given:

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

Calculate:

• $P(4)=(210)(0.53^4)(0.47^6)=0.1786117069\approx0.179$

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