Question

If the probability of success during a single event of a geometric experiment is 0.16, what is the probability of success by the 8th event? Round your answer to the nearest tenth of a percent.

A. 82.5%
B. 75.2%
C. 70.5%
D. 79.2%

Answer 1

This question is almost exactly the same as one of yours i answered the other day, just with different numbers, but same method: http://openstudy.com/users/briannageorge1#/updates/51a007b4e4b04449b221f739

Answer 2

so b?

Answer 3

I don't know... you didn't show any working so it looks like you just guessed.

Answer 4

no i was working on it.. .16^8 x .84?

Answer 5

or .84^8x.16

Answer 6

I think this question is actually a bit different, it's not asking for the probability ON the 8th event, they want the probability of it happening by the 8th event. Which looks like you need to add the probability of it happening on the first event, 2nd event, third,... up to the 8th event.

Answer 7

i added them up and got 1.28

Answer 8

What did you add up?

You need to add the prob on the first event, 2nd, et (note this is really a geometric series with a=0.16 and r=0.84):

1st: 0.16
2nd: 0.84*0.16
3rd: 0.84^2*0.16
4th: 0.84^3*0.16
5th: 0.84^4*0.16
6th: 0.84^5*0.16
7th: 0.84^6*0.16
8th: 0.84^7*0.16

Answer 9

.16 + 0.1344 + 0.112896 + 0.09483264 + 0.0796594176 + 0.06691391078 + 0.05620768505 + 0.04721445544

Answer 10

which equals 0.75212410887

Answer 11

so thats B!

Answer 12

thanks for your help

Answer 13

No prob. You could've also added it up using the geometric sum formula: http://fym.la.asu.edu/~tturner/MAT_117_online/SequenceAndSeries/geometric_sequence_18.gif

Answer 14

\[\Large S = 0.16\left( \frac{ 1- 0.84^8 }{ 1-0.84 } \right)\]

Answer 15

@agent0smith , so that means that this is true?

The expression P(1-P)^n, where P is the probability of success during a single event, represents the probability of success during the nth event of a geometric experiment.

Answer 16

Yes, that one tells you the probability of it happening on any single event. But are you sure it's not \[\Large P(1-P)^{n-1}\] since on the first event (n=1) the probability is just P.

Answer 17

My next question on my quiz is asking me that!

Answer 18

its true or false

Answer 19

http://isites.harvard.edu/fs/docs/icb.topic1197907.files/9-%20GeometricSeries.pdf

it has a n-1 (or k-1) here, as i suspected.

Answer 20

so false??

Answer 21

Yes, it's false, it would be \[\Large P(1-P)^{n-1}\]

Answer 22

you rock!

Answer 23

Thanks! :)