Answer :
Using the binomial distribution, it is found that:
- The probability that 6 VIPs attend is of 0.0055 = 0.55%.
- The probability that 10 VIPs attend is of approximately 0.
- The probability that more than 6 VIPs attend is 0.0009 = 0.09%.
What is the binomial distribution formula?
The formula is:
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]
The parameters are:
- x is the number of successes.
- n is the number of trials.
- p is the probability of a success on a single trial.
For this problem, the values of the parameters are:
n = 10, p = 0.2.
The probability that 6 VIPs attend is given by:
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 6) = C_{10,6}.(0.2)^{6}.(0.8)^{4} = 0.0055[/tex]
The probability that 10 VIPs attend is given by:
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 10) = C_{10,10}.(0.2)^{10}.(0.8)^{0} \approx 0[/tex]
The probability that more than 6 VIPs attend is given by:
P(X > 6) = P(X = 7) + P(X = 8) + P(X = 9) + P(X = 10).
Then:
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 7) = C_{10,7}.(0.2)^{7}.(0.8)^{3} = 0.0008[/tex]
[tex]P(X = 8) = C_{10,8}.(0.2)^{8}.(0.8)^{2} = 0.0001[/tex]
[tex]P(X = 9) = C_{10,9}.(0.2)^{9}.(0.8)^{1} \approx 0[/tex]
[tex]P(X = 10) = C_{10,10}.(0.2)^{10}.(0.8)^{0} \approx 0[/tex]
So:
P(X > 6) = P(X = 7) + P(X = 8) + P(X = 9) + P(X = 10) = 0.0008 + 0.0001 + 0 + 0 = 0.0009.
The probability that more than 6 VIPs attend is 0.0009 = 0.09%.
More can be learned about the binomial distribution at https://brainly.com/question/24863377
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