Unfair coin question The 2019 Stack Overflow Developer Survey Results Are InUnfair coin flippingGiven $5$ heads in $17$ coin flips of a biased coin, what is the probability at least $3$ of those heads occured in the first $10$ flips?If I flip a nonbiased coin 13 times what is the probability that I get tails 10 times?A coin is flipped 6 times. What is the probability that heads and tails occur an equal number of times?Probability of Unfair coin with p between (0, 1) and find # of heads s.t. it is even or |3.probability that we stop flipping after exactly ten flips in a biased coin flipping?What is the probability of heads in unfair coin when you flip the coin ten times?binomial distribution unfair coinFlip an unfair coin. Find: (a) p, exactly 7 heads (b) p, exactly 7 tails (c) p, atleast 7 headsCoin flipping and Bayes' theorem… but where does binomial theorem come in?
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Unfair coin question
The 2019 Stack Overflow Developer Survey Results Are InUnfair coin flippingGiven $5$ heads in $17$ coin flips of a biased coin, what is the probability at least $3$ of those heads occured in the first $10$ flips?If I flip a nonbiased coin 13 times what is the probability that I get tails 10 times?A coin is flipped 6 times. What is the probability that heads and tails occur an equal number of times?Probability of Unfair coin with p between (0, 1) and find # of heads s.t. it is even or |3.probability that we stop flipping after exactly ten flips in a biased coin flipping?What is the probability of heads in unfair coin when you flip the coin ten times?binomial distribution unfair coinFlip an unfair coin. Find: (a) p, exactly 7 heads (b) p, exactly 7 tails (c) p, atleast 7 headsCoin flipping and Bayes' theorem… but where does binomial theorem come in?
$begingroup$
Supposed that you flip an unfair coin with the probability of heads being $p(rm heads) = 3/4$ and the probability of tails being $1/4$ a total of ten times. How do you find the possibility of getting exactly $x$ heads and at least $x$ heads?
discrete-mathematics probability-distributions binomial-distribution
edited Apr 6 at 20:27
David G. Stork
12.2k41836
asked Apr 6 at 19:50
MarkMark
32
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Mark is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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add a comment |
$begingroup$
Supposed that you flip an unfair coin with the probability of heads being $p(rm heads) = 3/4$ and the probability of tails being $1/4$ a total of ten times. How do you find the possibility of getting exactly $x$ heads and at least $x$ heads?
discrete-mathematics probability-distributions binomial-distribution
edited Apr 6 at 20:27
David G. Stork
12.2k41836
asked Apr 6 at 19:50
MarkMark
32
New contributor
Mark is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
Supposed that you flip an unfair coin with the probability of heads being $p(rm heads) = 3/4$ and the probability of tails being $1/4$ a total of ten times. How do you find the possibility of getting exactly $x$ heads and at least $x$ heads?
discrete-mathematics probability-distributions binomial-distribution
edited Apr 6 at 20:27
David G. Stork
12.2k41836
asked Apr 6 at 19:50
MarkMark
32
New contributor
Mark is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
Supposed that you flip an unfair coin with the probability of heads being $p(rm heads) = 3/4$ and the probability of tails being $1/4$ a total of ten times. How do you find the possibility of getting exactly $x$ heads and at least $x$ heads?
discrete-mathematics probability-distributions binomial-distribution
discrete-mathematics probability-distributions binomial-distribution
edited Apr 6 at 20:27
David G. Stork
12.2k41836
asked Apr 6 at 19:50
MarkMark
32
New contributor
Mark is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited Apr 6 at 20:27
David G. Stork
12.2k41836
asked Apr 6 at 19:50
MarkMark
32
New contributor
Mark is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited Apr 6 at 20:27
David G. Stork
12.2k41836
edited Apr 6 at 20:27
David G. Stork
12.2k41836
edited Apr 6 at 20:27
David G. Stork
12.2k41836
12.2k41836
asked Apr 6 at 19:50
MarkMark
32
New contributor
Mark is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Apr 6 at 19:50
MarkMark
32
asked Apr 6 at 19:50
MarkMark
32
32
New contributor
Mark is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Mark is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Mark is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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add a comment |
add a comment |
3 Answers
3
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$begingroup$
In general, if the probability of heads on a given toss is $p$, and you toss $n$ times, the probability of getting exactly $x$ heads is
$$bbox[10px,#ffd]P(textexactly xtext heads)=binomnxp^x (1-p)^n-x,quad x=0,1,ldots,n.$$
To get the probability of getting at least $x$ heads, you sum up the probability of getting exactly $x$ heads, $x+1$ heads, ..., $n$ heads. So
$$bbox[10px,#ffd]P(textat least xtext heads)
=sum_k=x^nbinomnkp^k(1-p)^n-k.$$
If you want probabilities for tails instead of heads, replace every $p$ above with $1-p$. Note that $n=10$ and $p=3/4$ in your particular example.
By the way, the number of heads you get is said to follow a binomial distribution. See the Wikipedia page for more information.
answered Apr 6 at 19:55
Minus One-TwelfthMinus One-Twelfth
3,388413
$endgroup$
add a comment |
$begingroup$
Hint: use the binomial distribution. See, in particular, the examples on the Wikipedia page.
answered Apr 6 at 19:52
parsiadparsiad
18.8k32554
$endgroup$
add a comment |
$begingroup$
The probability of getting exactly $x$ tails is $binom 10 x left (frac 3 4 right )^x left ( frac 1 4 right )^10-x.$
The probability of getting at least $x$ many heads is $sumlimits_k=x^10 binom 10 k left (frac 3 4 right )^k left ( frac 1 4 right )^10-k.$
answered Apr 6 at 19:54
Dbchatto67Dbchatto67
2,753622
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
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votes
active
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votes
$begingroup$
In general, if the probability of heads on a given toss is $p$, and you toss $n$ times, the probability of getting exactly $x$ heads is
$$bbox[10px,#ffd]P(textexactly xtext heads)=binomnxp^x (1-p)^n-x,quad x=0,1,ldots,n.$$
To get the probability of getting at least $x$ heads, you sum up the probability of getting exactly $x$ heads, $x+1$ heads, ..., $n$ heads. So
$$bbox[10px,#ffd]P(textat least xtext heads)
=sum_k=x^nbinomnkp^k(1-p)^n-k.$$
If you want probabilities for tails instead of heads, replace every $p$ above with $1-p$. Note that $n=10$ and $p=3/4$ in your particular example.
By the way, the number of heads you get is said to follow a binomial distribution. See the Wikipedia page for more information.
answered Apr 6 at 19:55
Minus One-TwelfthMinus One-Twelfth
3,388413
$endgroup$
add a comment |
$begingroup$
In general, if the probability of heads on a given toss is $p$, and you toss $n$ times, the probability of getting exactly $x$ heads is
$$bbox[10px,#ffd]P(textexactly xtext heads)=binomnxp^x (1-p)^n-x,quad x=0,1,ldots,n.$$
To get the probability of getting at least $x$ heads, you sum up the probability of getting exactly $x$ heads, $x+1$ heads, ..., $n$ heads. So
$$bbox[10px,#ffd]P(textat least xtext heads)
=sum_k=x^nbinomnkp^k(1-p)^n-k.$$
If you want probabilities for tails instead of heads, replace every $p$ above with $1-p$. Note that $n=10$ and $p=3/4$ in your particular example.
By the way, the number of heads you get is said to follow a binomial distribution. See the Wikipedia page for more information.
answered Apr 6 at 19:55
Minus One-TwelfthMinus One-Twelfth
3,388413
$endgroup$
add a comment |
$begingroup$
In general, if the probability of heads on a given toss is $p$, and you toss $n$ times, the probability of getting exactly $x$ heads is
$$bbox[10px,#ffd]P(textexactly xtext heads)=binomnxp^x (1-p)^n-x,quad x=0,1,ldots,n.$$
To get the probability of getting at least $x$ heads, you sum up the probability of getting exactly $x$ heads, $x+1$ heads, ..., $n$ heads. So
$$bbox[10px,#ffd]P(textat least xtext heads)
=sum_k=x^nbinomnkp^k(1-p)^n-k.$$
If you want probabilities for tails instead of heads, replace every $p$ above with $1-p$. Note that $n=10$ and $p=3/4$ in your particular example.
By the way, the number of heads you get is said to follow a binomial distribution. See the Wikipedia page for more information.
answered Apr 6 at 19:55
Minus One-TwelfthMinus One-Twelfth
3,388413
$endgroup$
In general, if the probability of heads on a given toss is $p$, and you toss $n$ times, the probability of getting exactly $x$ heads is
$$bbox[10px,#ffd]P(textexactly xtext heads)=binomnxp^x (1-p)^n-x,quad x=0,1,ldots,n.$$
To get the probability of getting at least $x$ heads, you sum up the probability of getting exactly $x$ heads, $x+1$ heads, ..., $n$ heads. So
$$bbox[10px,#ffd]P(textat least xtext heads)
=sum_k=x^nbinomnkp^k(1-p)^n-k.$$
If you want probabilities for tails instead of heads, replace every $p$ above with $1-p$. Note that $n=10$ and $p=3/4$ in your particular example.
By the way, the number of heads you get is said to follow a binomial distribution. See the Wikipedia page for more information.
answered Apr 6 at 19:55
Minus One-TwelfthMinus One-Twelfth
3,388413
edited Apr 6 at 20:01
answered Apr 6 at 19:55
Minus One-TwelfthMinus One-Twelfth
3,388413
answered Apr 6 at 19:55
Minus One-TwelfthMinus One-Twelfth
3,388413
answered Apr 6 at 19:55
Minus One-TwelfthMinus One-Twelfth
3,388413
3,388413
add a comment |
add a comment |
$begingroup$
Hint: use the binomial distribution. See, in particular, the examples on the Wikipedia page.
answered Apr 6 at 19:52
parsiadparsiad
18.8k32554
$endgroup$
add a comment |
$begingroup$
Hint: use the binomial distribution. See, in particular, the examples on the Wikipedia page.
answered Apr 6 at 19:52
parsiadparsiad
18.8k32554
$endgroup$
add a comment |
$begingroup$
Hint: use the binomial distribution. See, in particular, the examples on the Wikipedia page.
answered Apr 6 at 19:52
parsiadparsiad
18.8k32554
$endgroup$
Hint: use the binomial distribution. See, in particular, the examples on the Wikipedia page.
answered Apr 6 at 19:52
parsiadparsiad
18.8k32554
answered Apr 6 at 19:52
parsiadparsiad
18.8k32554
answered Apr 6 at 19:52
parsiadparsiad
18.8k32554
answered Apr 6 at 19:52
parsiadparsiad
18.8k32554
18.8k32554
add a comment |
add a comment |
$begingroup$
The probability of getting exactly $x$ tails is $binom 10 x left (frac 3 4 right )^x left ( frac 1 4 right )^10-x.$
The probability of getting at least $x$ many heads is $sumlimits_k=x^10 binom 10 k left (frac 3 4 right )^k left ( frac 1 4 right )^10-k.$
answered Apr 6 at 19:54
Dbchatto67Dbchatto67
2,753622
$endgroup$
add a comment |
$begingroup$
The probability of getting exactly $x$ tails is $binom 10 x left (frac 3 4 right )^x left ( frac 1 4 right )^10-x.$
The probability of getting at least $x$ many heads is $sumlimits_k=x^10 binom 10 k left (frac 3 4 right )^k left ( frac 1 4 right )^10-k.$
answered Apr 6 at 19:54
Dbchatto67Dbchatto67
2,753622
$endgroup$
add a comment |
$begingroup$
The probability of getting exactly $x$ tails is $binom 10 x left (frac 3 4 right )^x left ( frac 1 4 right )^10-x.$
The probability of getting at least $x$ many heads is $sumlimits_k=x^10 binom 10 k left (frac 3 4 right )^k left ( frac 1 4 right )^10-k.$
answered Apr 6 at 19:54
Dbchatto67Dbchatto67
2,753622
$endgroup$
The probability of getting exactly $x$ tails is $binom 10 x left (frac 3 4 right )^x left ( frac 1 4 right )^10-x.$
The probability of getting at least $x$ many heads is $sumlimits_k=x^10 binom 10 k left (frac 3 4 right )^k left ( frac 1 4 right )^10-k.$
answered Apr 6 at 19:54
Dbchatto67Dbchatto67
2,753622
answered Apr 6 at 19:54
Dbchatto67Dbchatto67
2,753622
answered Apr 6 at 19:54
Dbchatto67Dbchatto67
2,753622
answered Apr 6 at 19:54
Dbchatto67Dbchatto67
2,753622
2,753622
add a comment |
add a comment |
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