finding the probability for coin toss and expected number of flips required, is this the right answer for...
$begingroup$
A coin having probability $p$ of coming up head is to be successively flipped until the first head appears.
- find the probability that the first head occurs in odd number of tosses ?
- find the expected number of flips required for the first head ?
the solution for number 1
$p(1 + (1-p)^2 + (1-p)^4 + ldots)
= pdisplaystylesum_{i=0}^infty ((1-p)^2)^i = frac{p}{1-(1-p)^2} = frac{1}{2-p}.$
the solution for number 2
$sumlimits_{i=1}^infty i p (1-p)^{i-1}=frac{1}{p}~~~~~~~$
is this the right answer for this problem and thank you
probability statistics probability-distributions
asked Dec 16 '18 at 19:03
green lifegreen life
186
$endgroup$
add a comment |
$begingroup$
A coin having probability $p$ of coming up head is to be successively flipped until the first head appears.
- find the probability that the first head occurs in odd number of tosses ?
- find the expected number of flips required for the first head ?
the solution for number 1
$p(1 + (1-p)^2 + (1-p)^4 + ldots)
= pdisplaystylesum_{i=0}^infty ((1-p)^2)^i = frac{p}{1-(1-p)^2} = frac{1}{2-p}.$
the solution for number 2
$sumlimits_{i=1}^infty i p (1-p)^{i-1}=frac{1}{p}~~~~~~~$
is this the right answer for this problem and thank you
probability statistics probability-distributions
asked Dec 16 '18 at 19:03
green lifegreen life
186
$endgroup$
add a comment |
$begingroup$
A coin having probability $p$ of coming up head is to be successively flipped until the first head appears.
- find the probability that the first head occurs in odd number of tosses ?
- find the expected number of flips required for the first head ?
the solution for number 1
$p(1 + (1-p)^2 + (1-p)^4 + ldots)
= pdisplaystylesum_{i=0}^infty ((1-p)^2)^i = frac{p}{1-(1-p)^2} = frac{1}{2-p}.$
the solution for number 2
$sumlimits_{i=1}^infty i p (1-p)^{i-1}=frac{1}{p}~~~~~~~$
is this the right answer for this problem and thank you
probability statistics probability-distributions
asked Dec 16 '18 at 19:03
green lifegreen life
186
$endgroup$
A coin having probability $p$ of coming up head is to be successively flipped until the first head appears.
- find the probability that the first head occurs in odd number of tosses ?
- find the expected number of flips required for the first head ?
the solution for number 1
$p(1 + (1-p)^2 + (1-p)^4 + ldots)
= pdisplaystylesum_{i=0}^infty ((1-p)^2)^i = frac{p}{1-(1-p)^2} = frac{1}{2-p}.$
the solution for number 2
$sumlimits_{i=1}^infty i p (1-p)^{i-1}=frac{1}{p}~~~~~~~$
is this the right answer for this problem and thank you
probability statistics probability-distributions
probability statistics probability-distributions
asked Dec 16 '18 at 19:03
green lifegreen life
186
asked Dec 16 '18 at 19:03
green lifegreen life
186
asked Dec 16 '18 at 19:03
green lifegreen life
186
asked Dec 16 '18 at 19:03
green lifegreen life
186
asked Dec 16 '18 at 19:03
green lifegreen life
186
186
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Both are correct.
As a way to avoid the use of infinite series:
For the first note that we either win on the first roll, lose on the second, or reset. Thus, letting $Psi$ denote the answer, we have $$Psi=p+(1-p)ptimes 0+(1-p)^2Psiimplies left(1-(1-p)^2right)Psi=pimplies Psi=frac p{1-(1-p^2)}$$
For the second note the recursive relation $$E=ptimes 1 +(1-p)times (E+1)implies pE=1implies E=frac 1p$$
answered Dec 16 '18 at 19:13
lulululu
42.9k25080
$endgroup$
$begingroup$
thank you very much for confirming the answer and for explaining this more clear way to obtain it
$endgroup$
– green life
Dec 16 '18 at 19:28
add a comment |
$begingroup$
Let $X = $ number of tosses until a head appears.
Then $X$ ~ Geometric$(frac{1}{2})$.
Part 1
The probability we wish to calculate is,
$$
Pr(odd) =
sum_{k=0}^{infty}{Pr(X = 2k + 1)} =
sum_{k=0}^{infty}{p(1 - p)^{2k}} =
psum_{k=0}^{infty}{((1-p)^{2})^{k}} =
frac{p}{1 - (1-p)^{2}} =
frac{1}{2-p}
$$
Part 2
Let $Y = $ heads was seen on the first flip.
Then $Y$ ~ Bernoulli$(p)$.
$$
E[X] = E[E[X|Y]] = Pr(Y = 1)E[X|Y=1] + Pr(Y=0)E[X| Y=0] = p(1) + (1-p)(1 + E[X])
$$
This then reduces to the equation,
$$
E[X] = 1 + (1-p)E[X]
$$
And then solving for $E[X]$ we get
$$
E[X] = frac{1}{p}
$$
Remarks
My calculations agree with yours, so I would conclude that your answers are correct.
answered Dec 16 '18 at 19:29
mwbrulhardtmwbrulhardt
11
$endgroup$
$begingroup$
thank you very much for confirming the answer and for the explanation
$endgroup$
– green life
Dec 16 '18 at 19:42
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Both are correct.
As a way to avoid the use of infinite series:
For the first note that we either win on the first roll, lose on the second, or reset. Thus, letting $Psi$ denote the answer, we have $$Psi=p+(1-p)ptimes 0+(1-p)^2Psiimplies left(1-(1-p)^2right)Psi=pimplies Psi=frac p{1-(1-p^2)}$$
For the second note the recursive relation $$E=ptimes 1 +(1-p)times (E+1)implies pE=1implies E=frac 1p$$
answered Dec 16 '18 at 19:13
lulululu
42.9k25080
$endgroup$
$begingroup$
thank you very much for confirming the answer and for explaining this more clear way to obtain it
$endgroup$
– green life
Dec 16 '18 at 19:28
add a comment |
$begingroup$
Both are correct.
As a way to avoid the use of infinite series:
For the first note that we either win on the first roll, lose on the second, or reset. Thus, letting $Psi$ denote the answer, we have $$Psi=p+(1-p)ptimes 0+(1-p)^2Psiimplies left(1-(1-p)^2right)Psi=pimplies Psi=frac p{1-(1-p^2)}$$
For the second note the recursive relation $$E=ptimes 1 +(1-p)times (E+1)implies pE=1implies E=frac 1p$$
answered Dec 16 '18 at 19:13
lulululu
42.9k25080
$endgroup$
$begingroup$
thank you very much for confirming the answer and for explaining this more clear way to obtain it
$endgroup$
– green life
Dec 16 '18 at 19:28
add a comment |
$begingroup$
Both are correct.
As a way to avoid the use of infinite series:
For the first note that we either win on the first roll, lose on the second, or reset. Thus, letting $Psi$ denote the answer, we have $$Psi=p+(1-p)ptimes 0+(1-p)^2Psiimplies left(1-(1-p)^2right)Psi=pimplies Psi=frac p{1-(1-p^2)}$$
For the second note the recursive relation $$E=ptimes 1 +(1-p)times (E+1)implies pE=1implies E=frac 1p$$
answered Dec 16 '18 at 19:13
lulululu
42.9k25080
$endgroup$
Both are correct.
As a way to avoid the use of infinite series:
For the first note that we either win on the first roll, lose on the second, or reset. Thus, letting $Psi$ denote the answer, we have $$Psi=p+(1-p)ptimes 0+(1-p)^2Psiimplies left(1-(1-p)^2right)Psi=pimplies Psi=frac p{1-(1-p^2)}$$
For the second note the recursive relation $$E=ptimes 1 +(1-p)times (E+1)implies pE=1implies E=frac 1p$$
answered Dec 16 '18 at 19:13
lulululu
42.9k25080
answered Dec 16 '18 at 19:13
lulululu
42.9k25080
answered Dec 16 '18 at 19:13
lulululu
42.9k25080
answered Dec 16 '18 at 19:13
lulululu
42.9k25080
42.9k25080
$begingroup$
thank you very much for confirming the answer and for explaining this more clear way to obtain it
$endgroup$
– green life
Dec 16 '18 at 19:28
add a comment |
$begingroup$
thank you very much for confirming the answer and for explaining this more clear way to obtain it
$endgroup$
– green life
Dec 16 '18 at 19:28
$begingroup$
thank you very much for confirming the answer and for explaining this more clear way to obtain it
$endgroup$
– green life
Dec 16 '18 at 19:28
$begingroup$
thank you very much for confirming the answer and for explaining this more clear way to obtain it
$endgroup$
– green life
Dec 16 '18 at 19:28
add a comment |
$begingroup$
Let $X = $ number of tosses until a head appears.
Then $X$ ~ Geometric$(frac{1}{2})$.
Part 1
The probability we wish to calculate is,
$$
Pr(odd) =
sum_{k=0}^{infty}{Pr(X = 2k + 1)} =
sum_{k=0}^{infty}{p(1 - p)^{2k}} =
psum_{k=0}^{infty}{((1-p)^{2})^{k}} =
frac{p}{1 - (1-p)^{2}} =
frac{1}{2-p}
$$
Part 2
Let $Y = $ heads was seen on the first flip.
Then $Y$ ~ Bernoulli$(p)$.
$$
E[X] = E[E[X|Y]] = Pr(Y = 1)E[X|Y=1] + Pr(Y=0)E[X| Y=0] = p(1) + (1-p)(1 + E[X])
$$
This then reduces to the equation,
$$
E[X] = 1 + (1-p)E[X]
$$
And then solving for $E[X]$ we get
$$
E[X] = frac{1}{p}
$$
Remarks
My calculations agree with yours, so I would conclude that your answers are correct.
answered Dec 16 '18 at 19:29
mwbrulhardtmwbrulhardt
11
$endgroup$
$begingroup$
thank you very much for confirming the answer and for the explanation
$endgroup$
– green life
Dec 16 '18 at 19:42
add a comment |
$begingroup$
Let $X = $ number of tosses until a head appears.
Then $X$ ~ Geometric$(frac{1}{2})$.
Part 1
The probability we wish to calculate is,
$$
Pr(odd) =
sum_{k=0}^{infty}{Pr(X = 2k + 1)} =
sum_{k=0}^{infty}{p(1 - p)^{2k}} =
psum_{k=0}^{infty}{((1-p)^{2})^{k}} =
frac{p}{1 - (1-p)^{2}} =
frac{1}{2-p}
$$
Part 2
Let $Y = $ heads was seen on the first flip.
Then $Y$ ~ Bernoulli$(p)$.
$$
E[X] = E[E[X|Y]] = Pr(Y = 1)E[X|Y=1] + Pr(Y=0)E[X| Y=0] = p(1) + (1-p)(1 + E[X])
$$
This then reduces to the equation,
$$
E[X] = 1 + (1-p)E[X]
$$
And then solving for $E[X]$ we get
$$
E[X] = frac{1}{p}
$$
Remarks
My calculations agree with yours, so I would conclude that your answers are correct.
answered Dec 16 '18 at 19:29
mwbrulhardtmwbrulhardt
11
$endgroup$
$begingroup$
thank you very much for confirming the answer and for the explanation
$endgroup$
– green life
Dec 16 '18 at 19:42
add a comment |
$begingroup$
Let $X = $ number of tosses until a head appears.
Then $X$ ~ Geometric$(frac{1}{2})$.
Part 1
The probability we wish to calculate is,
$$
Pr(odd) =
sum_{k=0}^{infty}{Pr(X = 2k + 1)} =
sum_{k=0}^{infty}{p(1 - p)^{2k}} =
psum_{k=0}^{infty}{((1-p)^{2})^{k}} =
frac{p}{1 - (1-p)^{2}} =
frac{1}{2-p}
$$
Part 2
Let $Y = $ heads was seen on the first flip.
Then $Y$ ~ Bernoulli$(p)$.
$$
E[X] = E[E[X|Y]] = Pr(Y = 1)E[X|Y=1] + Pr(Y=0)E[X| Y=0] = p(1) + (1-p)(1 + E[X])
$$
This then reduces to the equation,
$$
E[X] = 1 + (1-p)E[X]
$$
And then solving for $E[X]$ we get
$$
E[X] = frac{1}{p}
$$
Remarks
My calculations agree with yours, so I would conclude that your answers are correct.
answered Dec 16 '18 at 19:29
mwbrulhardtmwbrulhardt
11
$endgroup$
Let $X = $ number of tosses until a head appears.
Then $X$ ~ Geometric$(frac{1}{2})$.
Part 1
The probability we wish to calculate is,
$$
Pr(odd) =
sum_{k=0}^{infty}{Pr(X = 2k + 1)} =
sum_{k=0}^{infty}{p(1 - p)^{2k}} =
psum_{k=0}^{infty}{((1-p)^{2})^{k}} =
frac{p}{1 - (1-p)^{2}} =
frac{1}{2-p}
$$
Part 2
Let $Y = $ heads was seen on the first flip.
Then $Y$ ~ Bernoulli$(p)$.
$$
E[X] = E[E[X|Y]] = Pr(Y = 1)E[X|Y=1] + Pr(Y=0)E[X| Y=0] = p(1) + (1-p)(1 + E[X])
$$
This then reduces to the equation,
$$
E[X] = 1 + (1-p)E[X]
$$
And then solving for $E[X]$ we get
$$
E[X] = frac{1}{p}
$$
Remarks
My calculations agree with yours, so I would conclude that your answers are correct.
answered Dec 16 '18 at 19:29
mwbrulhardtmwbrulhardt
11
answered Dec 16 '18 at 19:29
mwbrulhardtmwbrulhardt
11
answered Dec 16 '18 at 19:29
mwbrulhardtmwbrulhardt
11
answered Dec 16 '18 at 19:29
mwbrulhardtmwbrulhardt
11
11
$begingroup$
thank you very much for confirming the answer and for the explanation
$endgroup$
– green life
Dec 16 '18 at 19:42
add a comment |
$begingroup$
thank you very much for confirming the answer and for the explanation
$endgroup$
– green life
Dec 16 '18 at 19:42
$begingroup$
thank you very much for confirming the answer and for the explanation
$endgroup$
– green life
Dec 16 '18 at 19:42
$begingroup$
thank you very much for confirming the answer and for the explanation
$endgroup$
– green life
Dec 16 '18 at 19:42
add a comment |
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