Question

On any given day there is a 15% chance that Ashley's internet connection will be lost at some point.
A. What is the probability that her service is uninterrupted for five days in a row?
B. What is the probability that her service is interrupted at least once in five days?

Answer 1

what are your thoughts about it?

Answer 2

i know i need to the probability

Answer 3

Ok lets have a look at the problem. The probability that the internet won't work today will be .15. Now, the probability that the internet still won't work tomorrow will be .15(day 1) * .15(day 2). So, for 5 days in a row, the prbability that it won't work will be (.15)^5

Answer 4

do you recall anything about say the binomial thrm?

Answer 5

Im not too sure about the binomial

Answer 6

its a nice form ... if i can recall how to code the latex for it.

\[P(x)=\binom{n}{x}p^x(1-p)^{n-x}\]
or something like that

Answer 7

if we determine our sample space of 5 days in a row and all the things that could occur:

qqqqq

pqqqq
qpqqq
qqpqq
qqqpq
qqqqp

ppqqq
pqpqq
....

we could then assign how many times our desired event occurs and the probability associated with it

Answer 8

in your case
\[\binom{5}{5}=1\]
is the number of ways we can choose 5 out of 5 ...
and our desired outcomes is working internet ... so 1 - .15

Answer 9

\[\binom{5}{5}.15^{(0)}(1-.15)^{(5)}\]
it fails 0 times, and works all 5 times

Answer 10

i get what your saying, but the numbers are losing me.

Answer 11

*you're

Is there an easier way than all of that? by in chance lol

Answer 12

i dont think there is an easier way to approach it.

consider this; the sum of all possible probabilities has to be 1: so we construct the binomial\[(p+(1-p))^n\]lets call 1-p=q to clean it up
\[(p+q)^n\]
if we expand the binomial we get a set of terms depicting the probabilties of each individual outcome
\[\binom{n}{0}p^nq^{0}+\binom{n}{1}p^{n-1}q^{1}+\binom{n}{2}p^{n-2}q^{2}+...\binom{n}{n}p^{n-n}q^{n}\]

Answer 13

if we know the expansion, we can determine the solutions quite readily yes?

Answer 14

our desired event has a: 1-.15=.85 probabilty of occuring

there is only 1 way in which all 5 days can work each time ...

1* (.85)^5

Answer 15

the other solution is similar, but requires the notion of a term i have forgotten since i took the classes :)

Answer 16

the concept being

P(at least 1) = 1 - P(none)

Answer 17

1 - (.15)^5

Answer 18

Ok, so 15% is interrupted internet and 85% is uninterrupted? so, (.85)^5= 0.4437

Answer 19

correct

Answer 20

so, to find the interrupted internet you take the AT LEAST which is.....

1- 0.4437?

Answer 21

yes ... i read the question a little off :) good thing you are paying attention tho

Answer 22

What is the probability that her service is interrupted at least once in five days?

1 - P(works all the time)

Answer 23

0.5563?

Answer 24

correct

Answer 25

ooooh... im shocked. yes! Thank you.

Answer 26

you did fine :) good luck