binomial distribution unfair coin
$begingroup$
I'm working on two bionmial distribution questions about unfair coin, where I got confused for one of em by its wording.
Unfair coin: You have a coin with which you are 2 times more likely to get heads than tails.
You flip the coin 100 times.
What is the probability of getting 20 tails? What is the probability of getting
at least one heads?
for 20 tails:
P(x=20) = 100!/(20!)(80!) * (1/3)^20 * (2/3)^80
anything wrong here?
for 'at least one head':
This is where I feel little lost.
do I get the sum of probability from getting 1 head to getting 100 heads? Is there any other trick to find the answer?
probability statistics binomial-distribution
asked Dec 19 '18 at 5:21
Daniel KimDaniel Kim
111
$endgroup$
add a comment |
$begingroup$
I'm working on two bionmial distribution questions about unfair coin, where I got confused for one of em by its wording.
Unfair coin: You have a coin with which you are 2 times more likely to get heads than tails.
You flip the coin 100 times.
What is the probability of getting 20 tails? What is the probability of getting
at least one heads?
for 20 tails:
P(x=20) = 100!/(20!)(80!) * (1/3)^20 * (2/3)^80
anything wrong here?
for 'at least one head':
This is where I feel little lost.
do I get the sum of probability from getting 1 head to getting 100 heads? Is there any other trick to find the answer?
probability statistics binomial-distribution
asked Dec 19 '18 at 5:21
Daniel KimDaniel Kim
111
$endgroup$
add a comment |
$begingroup$
I'm working on two bionmial distribution questions about unfair coin, where I got confused for one of em by its wording.
Unfair coin: You have a coin with which you are 2 times more likely to get heads than tails.
You flip the coin 100 times.
What is the probability of getting 20 tails? What is the probability of getting
at least one heads?
for 20 tails:
P(x=20) = 100!/(20!)(80!) * (1/3)^20 * (2/3)^80
anything wrong here?
for 'at least one head':
This is where I feel little lost.
do I get the sum of probability from getting 1 head to getting 100 heads? Is there any other trick to find the answer?
probability statistics binomial-distribution
asked Dec 19 '18 at 5:21
Daniel KimDaniel Kim
111
$endgroup$
I'm working on two bionmial distribution questions about unfair coin, where I got confused for one of em by its wording.
Unfair coin: You have a coin with which you are 2 times more likely to get heads than tails.
You flip the coin 100 times.
What is the probability of getting 20 tails? What is the probability of getting
at least one heads?
for 20 tails:
P(x=20) = 100!/(20!)(80!) * (1/3)^20 * (2/3)^80
anything wrong here?
for 'at least one head':
This is where I feel little lost.
do I get the sum of probability from getting 1 head to getting 100 heads? Is there any other trick to find the answer?
probability statistics binomial-distribution
probability statistics binomial-distribution
asked Dec 19 '18 at 5:21
Daniel KimDaniel Kim
111
asked Dec 19 '18 at 5:21
Daniel KimDaniel Kim
111
asked Dec 19 '18 at 5:21
Daniel KimDaniel Kim
111
asked Dec 19 '18 at 5:21
Daniel KimDaniel Kim
111
asked Dec 19 '18 at 5:21
Daniel KimDaniel Kim
111
111
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Your expression for $P(x = 20)$ looks good to me.
Hint For the other question: The complement of the event "at least one head" is the event "no heads".
answered Dec 19 '18 at 5:24
TravisTravis
60.6k767147
$endgroup$
$begingroup$
P(x=100) = 100!/100! * (1/3)^100 = 1/(3^100) = no heads. Therefore, 1 - 1/(3^100) ??
$endgroup$
– Daniel Kim
Dec 19 '18 at 5:27
$begingroup$
Looks good to me!
$endgroup$
– Travis
Dec 19 '18 at 5:45
$begingroup$
thank you Travis!
$endgroup$
– Daniel Kim
Dec 19 '18 at 6:22
$begingroup$
You're welcome, I'm glad you found it useful!
$endgroup$
– Travis
Dec 19 '18 at 6:43
$begingroup$
Travis, do you mind to check my question about exponential distribution as well? math.stackexchange.com/questions/3046085/…
$endgroup$
– Daniel Kim
Dec 19 '18 at 6:47
|
show 1 more comment
$begingroup$
What you can do is calculate the probability there is absolutely no head, then do one minus the probability of no head.
$$P(text{no head}) = left(frac13right)^{100},$$
so
$$P(text{at least 1 head}) = 1 - P(text{no head}).$$
edited Dec 19 '18 at 6:23
Saad
19.7k92352
answered Dec 19 '18 at 5:26
Muser_HaoMuser_Hao
142
$endgroup$
$begingroup$
thank you! I was also working on same thing via Travis's hint.
$endgroup$
– Daniel Kim
Dec 19 '18 at 5:28
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
$begingroup$
Your expression for $P(x = 20)$ looks good to me.
Hint For the other question: The complement of the event "at least one head" is the event "no heads".
answered Dec 19 '18 at 5:24
TravisTravis
60.6k767147
$endgroup$
$begingroup$
P(x=100) = 100!/100! * (1/3)^100 = 1/(3^100) = no heads. Therefore, 1 - 1/(3^100) ??
$endgroup$
– Daniel Kim
Dec 19 '18 at 5:27
$begingroup$
Looks good to me!
$endgroup$
– Travis
Dec 19 '18 at 5:45
$begingroup$
thank you Travis!
$endgroup$
– Daniel Kim
Dec 19 '18 at 6:22
$begingroup$
You're welcome, I'm glad you found it useful!
$endgroup$
– Travis
Dec 19 '18 at 6:43
$begingroup$
Travis, do you mind to check my question about exponential distribution as well? math.stackexchange.com/questions/3046085/…
$endgroup$
– Daniel Kim
Dec 19 '18 at 6:47
|
show 1 more comment
$begingroup$
Your expression for $P(x = 20)$ looks good to me.
Hint For the other question: The complement of the event "at least one head" is the event "no heads".
answered Dec 19 '18 at 5:24
TravisTravis
60.6k767147
$endgroup$
$begingroup$
P(x=100) = 100!/100! * (1/3)^100 = 1/(3^100) = no heads. Therefore, 1 - 1/(3^100) ??
$endgroup$
– Daniel Kim
Dec 19 '18 at 5:27
$begingroup$
Looks good to me!
$endgroup$
– Travis
Dec 19 '18 at 5:45
$begingroup$
thank you Travis!
$endgroup$
– Daniel Kim
Dec 19 '18 at 6:22
$begingroup$
You're welcome, I'm glad you found it useful!
$endgroup$
– Travis
Dec 19 '18 at 6:43
$begingroup$
Travis, do you mind to check my question about exponential distribution as well? math.stackexchange.com/questions/3046085/…
$endgroup$
– Daniel Kim
Dec 19 '18 at 6:47
|
show 1 more comment
$begingroup$
Your expression for $P(x = 20)$ looks good to me.
Hint For the other question: The complement of the event "at least one head" is the event "no heads".
answered Dec 19 '18 at 5:24
TravisTravis
60.6k767147
$endgroup$
Your expression for $P(x = 20)$ looks good to me.
Hint For the other question: The complement of the event "at least one head" is the event "no heads".
answered Dec 19 '18 at 5:24
TravisTravis
60.6k767147
answered Dec 19 '18 at 5:24
TravisTravis
60.6k767147
answered Dec 19 '18 at 5:24
TravisTravis
60.6k767147
answered Dec 19 '18 at 5:24
TravisTravis
60.6k767147
60.6k767147
$begingroup$
P(x=100) = 100!/100! * (1/3)^100 = 1/(3^100) = no heads. Therefore, 1 - 1/(3^100) ??
$endgroup$
– Daniel Kim
Dec 19 '18 at 5:27
$begingroup$
Looks good to me!
$endgroup$
– Travis
Dec 19 '18 at 5:45
$begingroup$
thank you Travis!
$endgroup$
– Daniel Kim
Dec 19 '18 at 6:22
$begingroup$
You're welcome, I'm glad you found it useful!
$endgroup$
– Travis
Dec 19 '18 at 6:43
$begingroup$
Travis, do you mind to check my question about exponential distribution as well? math.stackexchange.com/questions/3046085/…
$endgroup$
– Daniel Kim
Dec 19 '18 at 6:47
|
show 1 more comment
$begingroup$
P(x=100) = 100!/100! * (1/3)^100 = 1/(3^100) = no heads. Therefore, 1 - 1/(3^100) ??
$endgroup$
– Daniel Kim
Dec 19 '18 at 5:27
$begingroup$
Looks good to me!
$endgroup$
– Travis
Dec 19 '18 at 5:45
$begingroup$
thank you Travis!
$endgroup$
– Daniel Kim
Dec 19 '18 at 6:22
$begingroup$
You're welcome, I'm glad you found it useful!
$endgroup$
– Travis
Dec 19 '18 at 6:43
$begingroup$
Travis, do you mind to check my question about exponential distribution as well? math.stackexchange.com/questions/3046085/…
$endgroup$
– Daniel Kim
Dec 19 '18 at 6:47
$begingroup$
P(x=100) = 100!/100! * (1/3)^100 = 1/(3^100) = no heads. Therefore, 1 - 1/(3^100) ??
$endgroup$
– Daniel Kim
Dec 19 '18 at 5:27
$begingroup$
P(x=100) = 100!/100! * (1/3)^100 = 1/(3^100) = no heads. Therefore, 1 - 1/(3^100) ??
$endgroup$
– Daniel Kim
Dec 19 '18 at 5:27
$begingroup$
Looks good to me!
$endgroup$
– Travis
Dec 19 '18 at 5:45
$begingroup$
Looks good to me!
$endgroup$
– Travis
Dec 19 '18 at 5:45
$begingroup$
thank you Travis!
$endgroup$
– Daniel Kim
Dec 19 '18 at 6:22
$begingroup$
thank you Travis!
$endgroup$
– Daniel Kim
Dec 19 '18 at 6:22
$begingroup$
You're welcome, I'm glad you found it useful!
$endgroup$
– Travis
Dec 19 '18 at 6:43
$begingroup$
You're welcome, I'm glad you found it useful!
$endgroup$
– Travis
Dec 19 '18 at 6:43
$begingroup$
Travis, do you mind to check my question about exponential distribution as well? math.stackexchange.com/questions/3046085/…
$endgroup$
– Daniel Kim
Dec 19 '18 at 6:47
$begingroup$
Travis, do you mind to check my question about exponential distribution as well? math.stackexchange.com/questions/3046085/…
$endgroup$
– Daniel Kim
Dec 19 '18 at 6:47
|
show 1 more comment
$begingroup$
What you can do is calculate the probability there is absolutely no head, then do one minus the probability of no head.
$$P(text{no head}) = left(frac13right)^{100},$$
so
$$P(text{at least 1 head}) = 1 - P(text{no head}).$$
edited Dec 19 '18 at 6:23
Saad
19.7k92352
answered Dec 19 '18 at 5:26
Muser_HaoMuser_Hao
142
$endgroup$
$begingroup$
thank you! I was also working on same thing via Travis's hint.
$endgroup$
– Daniel Kim
Dec 19 '18 at 5:28
add a comment |
$begingroup$
What you can do is calculate the probability there is absolutely no head, then do one minus the probability of no head.
$$P(text{no head}) = left(frac13right)^{100},$$
so
$$P(text{at least 1 head}) = 1 - P(text{no head}).$$
edited Dec 19 '18 at 6:23
Saad
19.7k92352
answered Dec 19 '18 at 5:26
Muser_HaoMuser_Hao
142
$endgroup$
$begingroup$
thank you! I was also working on same thing via Travis's hint.
$endgroup$
– Daniel Kim
Dec 19 '18 at 5:28
add a comment |
$begingroup$
What you can do is calculate the probability there is absolutely no head, then do one minus the probability of no head.
$$P(text{no head}) = left(frac13right)^{100},$$
so
$$P(text{at least 1 head}) = 1 - P(text{no head}).$$
edited Dec 19 '18 at 6:23
Saad
19.7k92352
answered Dec 19 '18 at 5:26
Muser_HaoMuser_Hao
142
$endgroup$
What you can do is calculate the probability there is absolutely no head, then do one minus the probability of no head.
$$P(text{no head}) = left(frac13right)^{100},$$
so
$$P(text{at least 1 head}) = 1 - P(text{no head}).$$
edited Dec 19 '18 at 6:23
Saad
19.7k92352
answered Dec 19 '18 at 5:26
Muser_HaoMuser_Hao
142
edited Dec 19 '18 at 6:23
Saad
19.7k92352
edited Dec 19 '18 at 6:23
Saad
19.7k92352
edited Dec 19 '18 at 6:23
Saad
19.7k92352
19.7k92352
answered Dec 19 '18 at 5:26
Muser_HaoMuser_Hao
142
answered Dec 19 '18 at 5:26
Muser_HaoMuser_Hao
142
answered Dec 19 '18 at 5:26
Muser_HaoMuser_Hao
142
142
$begingroup$
thank you! I was also working on same thing via Travis's hint.
$endgroup$
– Daniel Kim
Dec 19 '18 at 5:28
add a comment |
$begingroup$
thank you! I was also working on same thing via Travis's hint.
$endgroup$
– Daniel Kim
Dec 19 '18 at 5:28
$begingroup$
thank you! I was also working on same thing via Travis's hint.
$endgroup$
– Daniel Kim
Dec 19 '18 at 5:28
$begingroup$
thank you! I was also working on same thing via Travis's hint.
$endgroup$
– Daniel Kim
Dec 19 '18 at 5:28
add a comment |
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