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Sagot :
Using the binomial distribution, it is found that there is a 0.0432 = 4.32% probability that she makes less than 6 of them.
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For each throw, there are only two possible outcomes. Either she makes it, or she misses it. The probability of making a throw is independent of any other throw, which means that the binomial probability distribution is used to solve this question.
Binomial probability distribution
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]
- x is the number of successes.
- n is the number of trials.
- p is the probability of a success on a single trial.
The probability of making less than 6 is:
[tex]P(X < 6) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5)[/tex]
In which
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 0) = C_{13,0}.(0.654)^{0}.(0.346)^{13} \approx 0[/tex]
[tex]P(X = 1) = C_{13,1}.(0.654)^{1}.(0.346)^{12} \approx 0[/tex]
[tex]P(X = 2) = C_{13,2}.(0.654)^{2}.(0.346)^{11} = 0.0003[/tex]
[tex]P(X = 3) = C_{13,3}.(0.654)^{3}.(0.346)^{10} = 0.0020[/tex]
[tex]P(X = 4) = C_{13,4}.(0.654)^{4}.(0.346)^{9} = 0.0093[/tex]
[tex]P(X = 5) = C_{13,5}.(0.654)^{5}.(0.346)^{8} = 0.0316[/tex]
Then
[tex]P(X < 6) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) = 0 + 0 + 0.0003 + 0.0020 + 0.0093 + 0.0316 = 0.0432[/tex]
0.0432 = 4.32% probability that she makes less than 6 of them.
A similar problem is given at https://brainly.com/question/24756209
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