Two types of coins are being tossed

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Let's assume we have two types of unfair coins. One has $70%$ probability of getting head, and the other has $70%$ change of getting tail. Now if we throw coin A $k$ times and coin B $n$ times. What will be the probability of getting at least $w$ heads. $w < k+n.$ I have it figured out when it comes to one coin. By using Normal distribution (aproximating Binomial distribution). But things get too complicated with second coin. Could you help me out?

probability statistics

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edited Jul 24 at 0:04

Michael Hardy

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asked Jul 22 at 15:48

John

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up vote
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Let's assume we have two types of unfair coins. One has $70%$ probability of getting head, and the other has $70%$ change of getting tail. Now if we throw coin A $k$ times and coin B $n$ times. What will be the probability of getting at least $w$ heads. $w < k+n.$ I have it figured out when it comes to one coin. By using Normal distribution (aproximating Binomial distribution). But things get too complicated with second coin. Could you help me out?

probability statistics

share|cite|improve this question

edited Jul 24 at 0:04

Michael Hardy

204k23186462

asked Jul 22 at 15:48

John

132

add a comment | 

up vote
2
down vote

favorite

up vote
2
down vote

favorite

Let's assume we have two types of unfair coins. One has $70%$ probability of getting head, and the other has $70%$ change of getting tail. Now if we throw coin A $k$ times and coin B $n$ times. What will be the probability of getting at least $w$ heads. $w < k+n.$ I have it figured out when it comes to one coin. By using Normal distribution (aproximating Binomial distribution). But things get too complicated with second coin. Could you help me out?

probability statistics

share|cite|improve this question

edited Jul 24 at 0:04

Michael Hardy

204k23186462

asked Jul 22 at 15:48

John

132

Let's assume we have two types of unfair coins. One has $70%$ probability of getting head, and the other has $70%$ change of getting tail. Now if we throw coin A $k$ times and coin B $n$ times. What will be the probability of getting at least $w$ heads. $w < k+n.$ I have it figured out when it comes to one coin. By using Normal distribution (aproximating Binomial distribution). But things get too complicated with second coin. Could you help me out?

probability statistics

share|cite|improve this question

edited Jul 24 at 0:04

Michael Hardy

204k23186462

asked Jul 22 at 15:48

John

132

share|cite|improve this question

share|cite|improve this question

edited Jul 24 at 0:04

Michael Hardy

204k23186462

edited Jul 24 at 0:04

Michael Hardy

204k23186462

edited Jul 24 at 0:04

Michael Hardy

204k23186462

204k23186462

asked Jul 22 at 15:48

John

132

asked Jul 22 at 15:48

John

132

asked Jul 22 at 15:48

John

132

132

add a comment | 

add a comment | 

1 Answer
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Let $X$ be the number of heads for coin $A$, and $Y$ the number for coin $B$. $X$ is approximately distributed like $N(0.7k, 0.21k)$, while $Y$ is approximately $N(0.3n, 0.21n)$. This means that $X+Y$ is approximately
$$
N(0.7k+0.3n,0.21(k+n)),
$$
so you can use a normal distribution with these parameters to approximate $X+Y$.

share|cite|improve this answer

answered Jul 22 at 16:05

Mike Earnest

15.2k11644

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Didn't think about summing the normal distribution. Looks good. Thank you :)
  – John
  Jul 22 at 16:12
  

  
  
  
  

add a comment | 

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1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
1
down vote



accepted

Let $X$ be the number of heads for coin $A$, and $Y$ the number for coin $B$. $X$ is approximately distributed like $N(0.7k, 0.21k)$, while $Y$ is approximately $N(0.3n, 0.21n)$. This means that $X+Y$ is approximately
$$
N(0.7k+0.3n,0.21(k+n)),
$$
so you can use a normal distribution with these parameters to approximate $X+Y$.

share|cite|improve this answer

answered Jul 22 at 16:05

Mike Earnest

15.2k11644

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Didn't think about summing the normal distribution. Looks good. Thank you :)
  – John
  Jul 22 at 16:12
  

  
  
  
  

add a comment | 

up vote
1
down vote



accepted

Let $X$ be the number of heads for coin $A$, and $Y$ the number for coin $B$. $X$ is approximately distributed like $N(0.7k, 0.21k)$, while $Y$ is approximately $N(0.3n, 0.21n)$. This means that $X+Y$ is approximately
$$
N(0.7k+0.3n,0.21(k+n)),
$$
so you can use a normal distribution with these parameters to approximate $X+Y$.

share|cite|improve this answer

answered Jul 22 at 16:05

Mike Earnest

15.2k11644

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Didn't think about summing the normal distribution. Looks good. Thank you :)
  – John
  Jul 22 at 16:12
  

  
  
  
  

add a comment | 

up vote
1
down vote



accepted

up vote
1
down vote



accepted

Let $X$ be the number of heads for coin $A$, and $Y$ the number for coin $B$. $X$ is approximately distributed like $N(0.7k, 0.21k)$, while $Y$ is approximately $N(0.3n, 0.21n)$. This means that $X+Y$ is approximately
$$
N(0.7k+0.3n,0.21(k+n)),
$$
so you can use a normal distribution with these parameters to approximate $X+Y$.

share|cite|improve this answer

answered Jul 22 at 16:05

Mike Earnest

15.2k11644

Let $X$ be the number of heads for coin $A$, and $Y$ the number for coin $B$. $X$ is approximately distributed like $N(0.7k, 0.21k)$, while $Y$ is approximately $N(0.3n, 0.21n)$. This means that $X+Y$ is approximately
$$
N(0.7k+0.3n,0.21(k+n)),
$$
so you can use a normal distribution with these parameters to approximate $X+Y$.

share|cite|improve this answer

answered Jul 22 at 16:05

Mike Earnest

15.2k11644

share|cite|improve this answer

share|cite|improve this answer

answered Jul 22 at 16:05

Mike Earnest

15.2k11644

answered Jul 22 at 16:05

Mike Earnest

15.2k11644

answered Jul 22 at 16:05

Mike Earnest

15.2k11644

15.2k11644

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Didn't think about summing the normal distribution. Looks good. Thank you :)
  – John
  Jul 22 at 16:12
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Didn't think about summing the normal distribution. Looks good. Thank you :)
  – John
  Jul 22 at 16:12
  

  
  
  
  

Didn't think about summing the normal distribution. Looks good. Thank you :)
– John
Jul 22 at 16:12

Didn't think about summing the normal distribution. Looks good. Thank you :)
– John
Jul 22 at 16:12

add a comment | 

 

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