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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
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