Question

A coin is unbalanced such that it comes up tails 55% of the time.

a) What is the probability that you don't flip a head until the 4th flip?
b) What is the probability that you don't flip a tail until the 4th flip?
c) How many tosses would it take, on average, to flip a head?
d) How many tosses would it take, on average, to flip a tail?

Answer 1

@robtobey what do you think

Answer 2

a.
\[(55/100)^3 (45/100) \]

Answer 3

for a i get .07486875

Answer 4

0.07487 is what I got after rounding.

Answer 5

okay great! wouldn't B be the same answer?

Answer 6

I think:
\[\large {\left( {\frac{{45}}{{100}}} \right)^3} \times \frac{{55}}{{100}}\]

Answer 7

so for b) .05011875?

Answer 8

yes!

Answer 9

what about c and d?

Answer 10

wait I'm confused how do we figure that then

Answer 11

I give you more explanation

Answer 12

we have a coin and the probability to get a tail is 55/100, whereas the probability to get a head is 45/100
so I call p=45/100 and q=1-p=55/100

Answer 13

Now the probability to get a tail after N tosses, is:
\[\large \left( {\begin{array}{*{20}{c}}
N \\
1
\end{array}} \right){\left( {\frac{{45}}{{100}}} \right)^{N - 1}}\frac{{55}}{{100}}\]

Answer 14

whereas the probability to get a head after N tosses, is:
\[\large \left( {\begin{array}{*{20}{c}}
N \\
1
\end{array}} \right){\left( {\frac{{55}}{{100}}} \right)^{N - 1}}\frac{{45}}{{100}}\]

Answer 15

okay

Answer 16

so, for answer c and d, we have to compute such N in order to get the requested probability

Answer 17

how do we get N

Answer 18

we can get N using the formulas above.

Answer 19

please wait, I'm confused, because, if I apply the binomial distribution, then the answers to questions a. and b. would be different by a factor which is a binomial coefficient

Answer 20

okay

Answer 21

I ask to another tutor, please wait...
@dan815 can you help here please?

Answer 22

@TuringTest can you help here please?

Answer 23

@amistre64 can you help here please?

Answer 24

You don't need the binomial coefficient

Answer 25

you would need the binomial coefficient if the question was "what are the odds of getting one heads/tails within the first 4 tosses"

Answer 26

ok!
what are the answers to questions c. and d. @TuringTest

Answer 27

hmm, if we wanted to know the number .. yeah

Answer 28

There is a nifty proof to the effect that a geometric rv with probability p has expected value of 1/p. I would be hard pressed to derive it off the top of my head though :P

Answer 29

P(1h in 4 tosses) = (4 1) ...

ut not getting it til the 4th toss is just: P(t)P(t)P(t)P(h)

Answer 30

How many tosses would it take, on average, to flip a head?

h = 45 out of 100

45/45 = 100/45 seems reasonable to me

Answer 31

so 45/45 is the answer?

Answer 32

of course 45/45 is not the answer ...

Answer 33

im so confused

Answer 34

so you are referring to the mean value
and to this equation:
1 = N*p
1= N*(45/100)?
@amistre64

Answer 35

45 tosses out of 100 tosses on average

what is 1 toss out of n tosses?

Answer 36

2.222222

Answer 37

I have thought to that formula, but I was not sure

Answer 38

i have no reference, just a little common sense is all

say its fair

50h out of 100 tosses

50/50 = 100/50

1h in 2 tosses

Answer 39

Here is proof that amistre is right
http://mathaa.epfl.ch/cours/PMMI2001/interactive/geomexpect_en0.htm

Answer 40

45 h out of 100 tosses, we woud expect to get 45/45 heads out of 100/45 tosses

Answer 41

ok! thank you! @amistre64 @TuringTest

Answer 42

since 2.22 is not a valid toss ... 3 tosses should suffice

Answer 43

okay so these are my answers:
a) .07486875
b).05011875
c) 2.222 rounded to three
d) 1.818 rounded to two

Answer 44

seems good to me yes

Answer 45

okay thank you all so much!

Answer 46

thank you all!

Answer 47

yalls welcome :)