Respuesta :
Answer:
0.999 is the required probability
Step-by-step explanation:
We are given the following information:
We treat graduates moving to a different state as a success.
P(graduates move to a different state) = 61% = 0.61
Then the number of adults follows a binomial distribution, where
[tex]P(X=x) = \binom{n}{x}.p^x.(1-p)^{n-x}[/tex]
where n is the total number of observations, x is the number of success, p is the probability of success.
Now, we are given n = 9
We have to evaluate:
[tex]P(x\geq 1)\\=1 = P(x = 0)\\=1-\binom{9}{0}(0.61)^0(1-0.61)^9\\=1-0.0002\\=0.999[/tex]
0.999 is the probability that among 9 randomly selected graduates, at least one moves to a different state after graduating.
Answer:
Probability that at least one moves to a different state after graduating is 0.999.
Step-by-step explanation:
We are given that a study conducted at a certain college shows that 61% of the school's graduates move to a different state after graduating.
Also, 9 graduates randomly selected and we have to find the probability that at least one moves to a different state after graduating.
The above situation can be represented through Binomial distribution;
[tex]P(X=r) = \binom{n}{r}p^{r} (1-p)^{n-r} ; x = 0,1,2,3,.....[/tex]
where, n = number of trials (samples) taken = 9 graduates
r = number of success = at least one
p = probability of success which in our question is % of school's
graduates moving to a different state after graduating, i.e; 61%
LET X = Number of graduates moving to a different state after graduating
So, it means X ~ [tex]Binom(n=9, p=0.61)[/tex]
Now, Probability that at least one moves to a different state after graduating is given by = P(X [tex]\geq[/tex] 1)
P(X [tex]\geq[/tex] 1) = 1 - P(X = 0)
= [tex]1-\binom{9}{0}\times 0.61^{0} \times (1-0.61)^{9-0}[/tex]
= [tex]1-(1 \times 1 \times 0.39^{9})[/tex]
= [tex]1-0.39^{9}[/tex] = 0.999
Therefore, Probability that among 9 randomly selected graduates, at least one moves to a different state after graduating is 0.999.
