Get the answers you need at Westonci.ca, where our expert community is always ready to help with accurate information. Discover a wealth of knowledge from professionals across various disciplines on our user-friendly Q&A platform. Discover detailed answers to your questions from a wide network of experts on our comprehensive Q&A platform.
Sagot :
Answer:
0.3758 = 37.58% probability that at most one of her first 10 arrows hits the target
Step-by-step explanation:
For each shot, there are only two possible outcomes. Either they hit the target, or they do not. The probability of a shot hitting the target is independent of any other shot, which means that the binomial probability distribution is used to solve this question.
Binomial probability distribution
The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
In which [tex]C_{n,x}[/tex] is the number of different combinations of x objects from a set of n elements, given by the following formula.
[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]
And p is the probability of X happening.
Each arrow with a probability 0.2
This means that [tex]p = 0.2[/tex]
First 10 arrows
This means that [tex]n = 10[/tex]
What is the probability that at most one of her first 10 arrows hits the target?
This is:
[tex]P(X \leq 1) = P(X = 0) + P(X = 1)[/tex]
So
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 0) = C_{10,0}.(0.2)^{0}.(0.8)^{10} = 0.1074[/tex]
[tex]P(X = 1) = C_{10,1}.(0.2)^{1}.(0.8)^{9} = 0.2684[/tex]
Then
[tex]P(X \leq 1) = P(X = 0) + P(X = 1) = 0.1074 + 0.2684 = 0.3758[/tex]
0.3758 = 37.58% probability that at most one of her first 10 arrows hits the target
We appreciate your time. Please come back anytime for the latest information and answers to your questions. Thanks for stopping by. We strive to provide the best answers for all your questions. See you again soon. We're here to help at Westonci.ca. Keep visiting for the best answers to your questions.
