Using the binomial distribution, it is found that you have a 39.6% probability of winning.
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For each trial, there are only two possible outcomes. Either you hit the target, or you do not. The probability of hitting the target on a trial is independent of any other trial, which means that the binomial distribution is used to solve this question.
Binomial probability distribution
It is the probability of exactly x successes on n repeated trials, with p probability, given by:
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]C_{n,x}[/tex] is the number of different combinations of x objects from a set of n elements, given by:
[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]
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- 63% probability of hitting the target, thus [tex]p = 0.63[/tex]
- Two throws, thus [tex]n = 2[/tex]
You have to hit it twice to win, thus the probability is P(X = 2).
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 2) = C_{2,2}.(0.63)^{2}.(0.37)^{0} = 0.396[/tex]
0.396 = 39.6% probability of winning.
A similar problem is given at https://brainly.com/question/23780714
