Question

For the Experiment, flip a coin until heads shows, assume that the probability of heads on one flip is 3/5. We define a RV (random variable) X= the number of flip.
Find the probability that X will be no more than 10. You may find it helpful to use the formula fo a partial sum of a geometry series, or you may us another method ( but you must explain or show you computed your result, as in all of these problems).

Answer 1

i guess you could compute the probability that it is greater than ten and subtract that from 1
not sure what is easiest though

Answer 2

in other words, without using the partial sum of a geometric sequence, not sure how you would do it

Answer 3

so i need the partial sum?

Answer 4

damn site keeps crashing on me

Answer 5

what i was trying to say was that it is worded in such a way that it suggests and easier method
but i cannot think of one off the top of my head

Answer 6

ok i see

Answer 7

should be easy enough to add though
btw did you get the other one?

Answer 8

the other one no lol

Answer 9

still thinking about it

Answer 10

we can do that if you like
this one should be very close to one

Answer 11

\[\frac{3}{5}\times\sum_{k=0}^{9}\left(\frac{2}{5}\right)^k\] for this one

Answer 12

the first one is actually easier
it is
\[\frac{3}{5}\sum_{k=0}^{\infty}\left(\frac{2}{5}\right)^k\]

Answer 13

k i see

Answer 14

if it is not clear how i got those, let me know

Answer 15

k=9 and infinity ?

Answer 16

the second one was for your previous problem
the 9 is for this problem

Answer 17

you mean the previous previous problem about prove ?

Answer 18

right that one

Answer 19

you had to prove that it added to one if i remember correctly

Answer 20

yes to prove that all probability is 1using the same geometry series

Answer 21

yes and you can sum
\[\sum_{k=0}^{\infty}(\frac{2}{5})^k\] in your head

Answer 22

you know how to do that?

Answer 23

or more to the point, is it clear how to get the thing you need to sum?

Answer 24

for this problem or the previous?

Answer 25

either
they are almost identical

Answer 26

i guess i need to \frac{ 3 }{ 5 } sum_{k=0}^{9}( frac{ 2 }{ 5 })^(9

Answer 27

\[\frac{ 3 }{ 5 } \sum_{k=0}^{9} (\frac{ 2 }{ 5 })^9\]

Answer 28

yeah
just asking if it is clear why that is the right thing
it is easy enough to sum

Answer 29

k so is (2/5)^9- (2/5)^0?

Answer 30

the sum will be
\[\frac{1-\left(\frac{2}{5}\right)^{10}}{1-\frac{2}{5}}\] then multiply by \(\frac{3}{5}\)
that sum is very close to \(\frac{5}{3}\) so your answer will be almost 1

Answer 31

since \((\frac{2}{5})^{10}\) is negligible

Answer 32

i see as a large numbers

Answer 33

not sure what you mean by large
it is very close to \(\frac{5}{3}\)
http://www.wolframalpha.com/input/?i=%281-%282%2F5%29^%2810%29%29%2F%283%2F5%29

Answer 34

\[(2/5)^9\]

Answer 35

that number is almost zero
in but in fact it should be
\[\frac{3}{5}\times\sum_{k=0}^{9}\left(\frac{2}{5}\right)^k=\frac{3}{5}\times \frac{1-\left(\frac{2}{5}\right)^{10}}{1-\frac{2}{5}}\]

Answer 36

is 3/5* (2/5)^9

Answer 37

?

Answer 38

no it is the one i wrote

Answer 39

first try prob is \(.6\)
second try it is \(.4\times .6\)
third try it is \(.4^2\times .6\)
fourth try it is \(.4^3\times .6\)
tenth try it is \(.4^9\times .6\) the sum is \[\frac{3}{5}\times\sum_{k=0}^{9}\left(\frac{2}{5}\right)^k=\frac{3}{5}\times \frac{1-\left(\frac{2}{5}\right)^{10}}{1-\frac{2}{5}}\]
\[

Answer 40

ok i see

Answer 41

k good
and as i said, that number is so close to one as to be essentially 1
which makes sense since if you flip a coin it is almost certain you will get a head by the tenth flip
even more so with this bias coin that favors heads

Answer 42

second .24, .96 and the .4^2 is larger numbers

Answer 43

is like 1.5728...

Answer 44

i mean the last one .4^9*.6