Evaluate 12sigma n=3 20(0.5)^n-1

1 answer:

5 0

For the same reasons as the last problem...

s(n)=20(1-(1/2)^n)/(1-1/2)

s(n)=40(1-(1/2)^n)

But this time we are not starting at one so we must subtract the sum of s(2) from the sum of s(12) so:

40((1-(1/2)^12-(1-(1/2)^2)

40(0.249755859375)

≈9.99 (to nearest one-thousandth)

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Step-by-step explanation:

We use the conditional probability formula to solve this question. It is

$P(B|A) = \frac{P(A \cap B)}{P(A)}$

In which

P(B|A) is the probability of event B happening, given that A happened.

$P(A \cap B)$ is the probability of both A and B happening.

P(A) is the probability of A happening.

Compute the probability of a Madonna University nursing student getting hired within 6 months of graduation given that they complete their degree here.

In this problem, we have that:

Event A: Completing their degree there.

There is a 34.72% chance of a student interested in a nursing degree not making it through the program.

So 100 - 34.72% = 65.28% = 0.6528 probability of completing their degree. So P(A) = 0.6528.

Event B: Getting hired

86.74% chance of getting hired full time in a hospital setting within 6 months of graduation.

Completing their degree and getting hired:

$P(A \cap B) = 0.6528*0.8674$