Probability of getting at least one head given there at least two heads. (solution verification)
Clash Royale CLAN TAG#URR8PPP
up vote
3
down vote
favorite
A coin with probability, $p$, of coming up heads is flipped three times (the flips are independent).
What is the probability of getting at least one head given there are at least two heads?
My attempt at a solution:
First I realize that there are eight possible outcomes: TTT, HHH, THH, HHT, HTH, TTH, THT, HTT. Now consider:
beginalign*
&P(1~texthead mid textat least $2$ heads) + P(2~textheads mid textat least $2$ heads)+ P(3~textheads mid textat least $2$ heads)\
& quad = fracP(1~texthead) cap P(textat least $2$ heads)P(textat least $2$ heads) + fracP(2~textheads) cap P(textat least $2$ heads)P(textat least $2$ heads)\
& qquad + fracP(3~textheads) cap P(textat least $2$ heads)P(textat least $2$ heads)\
& quad = 0 + fracP(2~textheads)P(textat least $2$ heads) + fracP(3~textheads)P(textat least $2$ heads)\
& quad = frac3p^2(1-p)3p^2(1-p)+ p^3 + fracp^33p^2(1-p)+ p^3
endalign*
Is this solution correct? I'm not sure whether the probability of getting two heads is $3p^2(1-p)$ or if it is $(p^2)(1-p)$. I figured that since there was three possible ways of getting two heads I would multiply by $3$ (add three times). But I'm not sure if this correct.
probability algebra-precalculus probability-theory statistics
edited Sep 4 at 8:43
N. F. Taussig
39.6k93153
asked Sep 4 at 7:14
bob bob
162
add a comment |Â
up vote
3
down vote
favorite
A coin with probability, $p$, of coming up heads is flipped three times (the flips are independent).
What is the probability of getting at least one head given there are at least two heads?
My attempt at a solution:
First I realize that there are eight possible outcomes: TTT, HHH, THH, HHT, HTH, TTH, THT, HTT. Now consider:
beginalign*
&P(1~texthead mid textat least $2$ heads) + P(2~textheads mid textat least $2$ heads)+ P(3~textheads mid textat least $2$ heads)\
& quad = fracP(1~texthead) cap P(textat least $2$ heads)P(textat least $2$ heads) + fracP(2~textheads) cap P(textat least $2$ heads)P(textat least $2$ heads)\
& qquad + fracP(3~textheads) cap P(textat least $2$ heads)P(textat least $2$ heads)\
& quad = 0 + fracP(2~textheads)P(textat least $2$ heads) + fracP(3~textheads)P(textat least $2$ heads)\
& quad = frac3p^2(1-p)3p^2(1-p)+ p^3 + fracp^33p^2(1-p)+ p^3
endalign*
Is this solution correct? I'm not sure whether the probability of getting two heads is $3p^2(1-p)$ or if it is $(p^2)(1-p)$. I figured that since there was three possible ways of getting two heads I would multiply by $3$ (add three times). But I'm not sure if this correct.
probability algebra-precalculus probability-theory statistics
edited Sep 4 at 8:43
N. F. Taussig
39.6k93153
asked Sep 4 at 7:14
bob bob
162
Welcome to MathSE. Please read this tutorial on how to typeset mathematics on this site.
– N. F. Taussig
Sep 4 at 8:19
@N.F.Taussig Hi, I was struggling typsetting in Latex or MathJax fractions with words in it. Should i use mbox?
– bob bob
Sep 4 at 8:24
A simpler solution is possible. Let $A$ be the event of getting at least one heads, and $B$ the event of getting at least two heads. Notice that $A cap B = B$. It follows that $P(A|B) = P(Acap B)/P(B) = P(B)/P(B)=1$.
– littleO
Sep 4 at 8:45
I have edited your equations using the align environment. I introduced the intersection symbol, which can be typeset using cap when you are in math mode. I also moved the text inside the fractions. See if you are happy with the results. If you right-click on the equations, you will see a box that says Show Math As at the top. If you click this, you can select the option TeX commands. Alternatively, you can click on the edit button to see what I did or improve on it.
– N. F. Taussig
Sep 4 at 8:48
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
A coin with probability, $p$, of coming up heads is flipped three times (the flips are independent).
What is the probability of getting at least one head given there are at least two heads?
My attempt at a solution:
First I realize that there are eight possible outcomes: TTT, HHH, THH, HHT, HTH, TTH, THT, HTT. Now consider:
beginalign*
&P(1~texthead mid textat least $2$ heads) + P(2~textheads mid textat least $2$ heads)+ P(3~textheads mid textat least $2$ heads)\
& quad = fracP(1~texthead) cap P(textat least $2$ heads)P(textat least $2$ heads) + fracP(2~textheads) cap P(textat least $2$ heads)P(textat least $2$ heads)\
& qquad + fracP(3~textheads) cap P(textat least $2$ heads)P(textat least $2$ heads)\
& quad = 0 + fracP(2~textheads)P(textat least $2$ heads) + fracP(3~textheads)P(textat least $2$ heads)\
& quad = frac3p^2(1-p)3p^2(1-p)+ p^3 + fracp^33p^2(1-p)+ p^3
endalign*
Is this solution correct? I'm not sure whether the probability of getting two heads is $3p^2(1-p)$ or if it is $(p^2)(1-p)$. I figured that since there was three possible ways of getting two heads I would multiply by $3$ (add three times). But I'm not sure if this correct.
probability algebra-precalculus probability-theory statistics
edited Sep 4 at 8:43
N. F. Taussig
39.6k93153
asked Sep 4 at 7:14
bob bob
162
A coin with probability, $p$, of coming up heads is flipped three times (the flips are independent).
What is the probability of getting at least one head given there are at least two heads?
My attempt at a solution:
First I realize that there are eight possible outcomes: TTT, HHH, THH, HHT, HTH, TTH, THT, HTT. Now consider:
beginalign*
&P(1~texthead mid textat least $2$ heads) + P(2~textheads mid textat least $2$ heads)+ P(3~textheads mid textat least $2$ heads)\
& quad = fracP(1~texthead) cap P(textat least $2$ heads)P(textat least $2$ heads) + fracP(2~textheads) cap P(textat least $2$ heads)P(textat least $2$ heads)\
& qquad + fracP(3~textheads) cap P(textat least $2$ heads)P(textat least $2$ heads)\
& quad = 0 + fracP(2~textheads)P(textat least $2$ heads) + fracP(3~textheads)P(textat least $2$ heads)\
& quad = frac3p^2(1-p)3p^2(1-p)+ p^3 + fracp^33p^2(1-p)+ p^3
endalign*
Is this solution correct? I'm not sure whether the probability of getting two heads is $3p^2(1-p)$ or if it is $(p^2)(1-p)$. I figured that since there was three possible ways of getting two heads I would multiply by $3$ (add three times). But I'm not sure if this correct.
probability algebra-precalculus probability-theory statistics
probability algebra-precalculus probability-theory statistics
edited Sep 4 at 8:43
N. F. Taussig
39.6k93153
asked Sep 4 at 7:14
bob bob
162
edited Sep 4 at 8:43
N. F. Taussig
39.6k93153
asked Sep 4 at 7:14
bob bob
162
edited Sep 4 at 8:43
N. F. Taussig
39.6k93153
edited Sep 4 at 8:43
N. F. Taussig
39.6k93153
edited Sep 4 at 8:43
N. F. Taussig
39.6k93153
39.6k93153
asked Sep 4 at 7:14
bob bob
162
asked Sep 4 at 7:14
bob bob
162
asked Sep 4 at 7:14
bob bob
162
162
Welcome to MathSE. Please read this tutorial on how to typeset mathematics on this site.
– N. F. Taussig
Sep 4 at 8:19
@N.F.Taussig Hi, I was struggling typsetting in Latex or MathJax fractions with words in it. Should i use mbox?
– bob bob
Sep 4 at 8:24
A simpler solution is possible. Let $A$ be the event of getting at least one heads, and $B$ the event of getting at least two heads. Notice that $A cap B = B$. It follows that $P(A|B) = P(Acap B)/P(B) = P(B)/P(B)=1$.
– littleO
Sep 4 at 8:45
I have edited your equations using the align environment. I introduced the intersection symbol, which can be typeset using cap when you are in math mode. I also moved the text inside the fractions. See if you are happy with the results. If you right-click on the equations, you will see a box that says Show Math As at the top. If you click this, you can select the option TeX commands. Alternatively, you can click on the edit button to see what I did or improve on it.
– N. F. Taussig
Sep 4 at 8:48
add a comment |Â
Welcome to MathSE. Please read this tutorial on how to typeset mathematics on this site.
– N. F. Taussig
Sep 4 at 8:19
@N.F.Taussig Hi, I was struggling typsetting in Latex or MathJax fractions with words in it. Should i use mbox?
– bob bob
Sep 4 at 8:24
A simpler solution is possible. Let $A$ be the event of getting at least one heads, and $B$ the event of getting at least two heads. Notice that $A cap B = B$. It follows that $P(A|B) = P(Acap B)/P(B) = P(B)/P(B)=1$.
– littleO
Sep 4 at 8:45
I have edited your equations using the align environment. I introduced the intersection symbol, which can be typeset using cap when you are in math mode. I also moved the text inside the fractions. See if you are happy with the results. If you right-click on the equations, you will see a box that says Show Math As at the top. If you click this, you can select the option TeX commands. Alternatively, you can click on the edit button to see what I did or improve on it.
– N. F. Taussig
Sep 4 at 8:48
Welcome to MathSE. Please read this tutorial on how to typeset mathematics on this site.
– N. F. Taussig
Sep 4 at 8:19
Welcome to MathSE. Please read this tutorial on how to typeset mathematics on this site.
– N. F. Taussig
Sep 4 at 8:19
@N.F.Taussig Hi, I was struggling typsetting in Latex or MathJax fractions with words in it. Should i use mbox?
– bob bob
Sep 4 at 8:24
@N.F.Taussig Hi, I was struggling typsetting in Latex or MathJax fractions with words in it. Should i use mbox?
– bob bob
Sep 4 at 8:24
A simpler solution is possible. Let $A$ be the event of getting at least one heads, and $B$ the event of getting at least two heads. Notice that $A cap B = B$. It follows that $P(A|B) = P(Acap B)/P(B) = P(B)/P(B)=1$.
– littleO
Sep 4 at 8:45
A simpler solution is possible. Let $A$ be the event of getting at least one heads, and $B$ the event of getting at least two heads. Notice that $A cap B = B$. It follows that $P(A|B) = P(Acap B)/P(B) = P(B)/P(B)=1$.
– littleO
Sep 4 at 8:45
I have edited your equations using the align environment. I introduced the intersection symbol, which can be typeset using cap when you are in math mode. I also moved the text inside the fractions. See if you are happy with the results. If you right-click on the equations, you will see a box that says Show Math As at the top. If you click this, you can select the option TeX commands. Alternatively, you can click on the edit button to see what I did or improve on it.
– N. F. Taussig
Sep 4 at 8:48
I have edited your equations using the align environment. I introduced the intersection symbol, which can be typeset using cap when you are in math mode. I also moved the text inside the fractions. See if you are happy with the results. If you right-click on the equations, you will see a box that says Show Math As at the top. If you click this, you can select the option TeX commands. Alternatively, you can click on the edit button to see what I did or improve on it.
– N. F. Taussig
Sep 4 at 8:48
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
3
down vote
If you have at least two heads then you definitely have at least one head. So you would have thought a probability of $1$ and indeed:
$$0+frac3p^2(1-p)3p^2(1-p)+ p^3+ fracp^33p^2(1-p)+ p^3 = 1$$
The probability of exactly $k$ heads from $n$ attempts is $n choose kp^k(1-p)^n-k$
which with $n=3$ and $k=2$ is equal to $3p^2(1-p)$
answered Sep 4 at 7:31
Henry
94.2k472150
thank you so my answer is correct?
– bob bob
Sep 4 at 8:11
@bobbob It would have been better if you had simplified your final expression to $1$
– Henry
Sep 4 at 9:50
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
If you have at least two heads then you definitely have at least one head. So you would have thought a probability of $1$ and indeed:
$$0+frac3p^2(1-p)3p^2(1-p)+ p^3+ fracp^33p^2(1-p)+ p^3 = 1$$
The probability of exactly $k$ heads from $n$ attempts is $n choose kp^k(1-p)^n-k$
which with $n=3$ and $k=2$ is equal to $3p^2(1-p)$
answered Sep 4 at 7:31
Henry
94.2k472150
thank you so my answer is correct?
– bob bob
Sep 4 at 8:11
@bobbob It would have been better if you had simplified your final expression to $1$
– Henry
Sep 4 at 9:50
add a comment |Â
up vote
3
down vote
If you have at least two heads then you definitely have at least one head. So you would have thought a probability of $1$ and indeed:
$$0+frac3p^2(1-p)3p^2(1-p)+ p^3+ fracp^33p^2(1-p)+ p^3 = 1$$
The probability of exactly $k$ heads from $n$ attempts is $n choose kp^k(1-p)^n-k$
which with $n=3$ and $k=2$ is equal to $3p^2(1-p)$
answered Sep 4 at 7:31
Henry
94.2k472150
thank you so my answer is correct?
– bob bob
Sep 4 at 8:11
@bobbob It would have been better if you had simplified your final expression to $1$
– Henry
Sep 4 at 9:50
add a comment |Â
up vote
3
down vote
up vote
3
down vote
If you have at least two heads then you definitely have at least one head. So you would have thought a probability of $1$ and indeed:
$$0+frac3p^2(1-p)3p^2(1-p)+ p^3+ fracp^33p^2(1-p)+ p^3 = 1$$
The probability of exactly $k$ heads from $n$ attempts is $n choose kp^k(1-p)^n-k$
which with $n=3$ and $k=2$ is equal to $3p^2(1-p)$
answered Sep 4 at 7:31
Henry
94.2k472150
If you have at least two heads then you definitely have at least one head. So you would have thought a probability of $1$ and indeed:
$$0+frac3p^2(1-p)3p^2(1-p)+ p^3+ fracp^33p^2(1-p)+ p^3 = 1$$
The probability of exactly $k$ heads from $n$ attempts is $n choose kp^k(1-p)^n-k$
which with $n=3$ and $k=2$ is equal to $3p^2(1-p)$
answered Sep 4 at 7:31
Henry
94.2k472150
answered Sep 4 at 7:31
Henry
94.2k472150
answered Sep 4 at 7:31
Henry
94.2k472150
answered Sep 4 at 7:31
Henry
94.2k472150
94.2k472150
thank you so my answer is correct?
– bob bob
Sep 4 at 8:11
@bobbob It would have been better if you had simplified your final expression to $1$
– Henry
Sep 4 at 9:50
add a comment |Â
thank you so my answer is correct?
– bob bob
Sep 4 at 8:11
@bobbob It would have been better if you had simplified your final expression to $1$
– Henry
Sep 4 at 9:50
thank you so my answer is correct?
– bob bob
Sep 4 at 8:11
thank you so my answer is correct?
– bob bob
Sep 4 at 8:11
@bobbob It would have been better if you had simplified your final expression to $1$
– Henry
Sep 4 at 9:50
@bobbob It would have been better if you had simplified your final expression to $1$
– Henry
Sep 4 at 9:50
add a comment |Â
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2904722%2fprobability-of-getting-at-least-one-head-given-there-at-least-two-heads-soluti%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Welcome to MathSE. Please read this tutorial on how to typeset mathematics on this site.
– N. F. Taussig
Sep 4 at 8:19
@N.F.Taussig Hi, I was struggling typsetting in Latex or MathJax fractions with words in it. Should i use mbox?
– bob bob
Sep 4 at 8:24
A simpler solution is possible. Let $A$ be the event of getting at least one heads, and $B$ the event of getting at least two heads. Notice that $A cap B = B$. It follows that $P(A|B) = P(Acap B)/P(B) = P(B)/P(B)=1$.
– littleO
Sep 4 at 8:45
I have edited your equations using the align environment. I introduced the intersection symbol, which can be typeset using cap when you are in math mode. I also moved the text inside the fractions. See if you are happy with the results. If you right-click on the equations, you will see a box that says Show Math As at the top. If you click this, you can select the option TeX commands. Alternatively, you can click on the edit button to see what I did or improve on it.
– N. F. Taussig
Sep 4 at 8:48