Mathematics, 30.03.2020 22:14 ferg6

A study conducted at a certain college shows that 61% of the school's graduates move to a different state after graduating. Find the probability that among 9 randomly selected graduates, at least one moves to a different state after graduating. Round to three decimal places as needed.

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Answer from: mega29

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

$P(X=x) = \binom{n}{x}.p^x.(1-p)^{n-x}$

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:

$P(x\geq 1)\\=1 = P(x = 0)\\=1-\binom{9}{0}(0.61)^0(1-0.61)^9\\=1-0.0002\\=0.999$

0.999 is the probability that among 9 randomly selected graduates, at least one moves to a different state after graduating.

Answer from: eylinglez3ovm16v

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;

$P(X=r) = \binom{n}{r}p^{r} (1-p)^{n-r} ; x = 0,1,2,3,.....$

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 ~ Binom(n=9, p=0.61)

Now, Probability that at least one moves to a different state after graduating is given by = P(X $\geq$ 1)

P(X $\geq$ 1) = 1 - P(X = 0)

= $1-\binom{9}{0}\times 0.61^{0} \times (1-0.61)^{9-0}$

= $1-(1 \times 1 \times 0.39^{9})$

= $1-0.39^{9}$ = 0.999

Therefore, Probability that among 9 randomly selected graduates, at least one moves to a different state after graduating is 0.999.

Answer from: Quest

hello from mrbilldoesmath!

answer:

a(n) - a(n-1) -3

discussion:

this ones a bit interesting. given a(n) = 5 - 3n consider this:

a(n-1) = 5 - 3(n-1)

= (5 - 3n) + 3 => but a(n) = 5 - 3n, the first term

= a(n) + 3 => subtract 3 from both sides

a(n-1) -3 = a(n)

a(n) - a(n-1) -3

you,

mrb

Answer from: Quest

idc

step-by-step explanation:

idk

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