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Six coins are tossed. Alice pays Betty…

Dice probability of exact number of headsConfused about fair coins?Six identical coins in a box, five unbiased and one biasedTwo coins tossed $5$ times with different probabilites for heads on each coin. What's the probability of getting at least $3$ heads?Probability of getting an odd number of heads if n biased coins are tossed once.Probability of getting a certain number of heads in coin toss problem with differently biased coins with different numbersA unfair coin is tossed until $A$ heads or $A$ tails are observedFinding the PMF of number of heads resulting from second round of tosses of coinsProbability of choosing a biased coin $C$ which has probability $3/15$ of getting heads, assuming we got head on the first tossWhat's the probability that picking $3$ coins from a bag containing all coins from $1$ cent through $2$ euro will yield a total of $80$ cents?

1

1

$begingroup$

Six coins are tossed. Alice pays Betty according to the following table:

If no heads: $200

If 1 head: $50

If 2 heads: $10

If 3 heads: $5

If 4 heads: $20

If 5 heads: $25

If 6 heads: $80

If $X$ is the amount Alice pays, find $E(X)$

I have no idea how to start this

EDIT:

Okay so for 0 heads, we have $binom60 * (frac12)^0 * (frac12)^6 = frac164$

1 heads: $binom61 * (frac12)^1 * (frac12)^5 = frac332$

2 heads: $binom62 * (frac12)^2 * (frac12)^4= frac1564$

3 heads: $binom63 * (frac12)^3 * (frac12)^3 = frac516$

4 heads: $binom64 * (frac12)^4 * (frac12)^2 = frac1564$

5 heads: $binom65 * (frac12)^5 * (frac12)^1 = frac332$

6 heads: $binom66 * (frac12)^6 * (frac12)^0 = frac164$

Then I take each of these values and multiply it with X right? So for example, for 0 heads which is $frac164$, I multiply it with 200, then add it with $frac332 * 50$ for 1 heads, etc.?

probability discrete-mathematics

share|cite|improve this question

edited 2 days ago

Stuy

asked 2 days ago

StuyStuy

4351412

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Are you able to calculate the probability for no head, 1 head, 2 heads, ...? Key word: Binomial distribution. Now it´s your turn!
  $endgroup$
  – callculus
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Well the thing I'm confused about is whether or not we have bernoulli trials. It says we toss 6 coins, rather than 1 coin 6 times in a row, so we're just tossing a total of 6 coins in one go. Each coin has a 1/2 chance of being heads or tails, and since there are 6 coins, then the probability is $frac12^6$?
  $endgroup$
  – Stuy
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes, tossing the coins one after another is the same as tossing them all at the same time.
  $endgroup$
  – saulspatz
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Stuy Yes, but this is only a part of the binomial distribution. You forgot the binomial coefficient. What is the prob. that $1$ coin is head?
  $endgroup$
  – callculus
  2 days ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Oh okay, so since we do have Bernoulli Trials, for 0 heads it would be $binom60 * (frac12)^0*(frac12)^5$?
  $endgroup$
  – Stuy
  2 days ago
  

  
  
  
  

 | 
show 4 more comments

1

1

$begingroup$

Six coins are tossed. Alice pays Betty according to the following table:

If no heads: $200

If 1 head: $50

If 2 heads: $10

If 3 heads: $5

If 4 heads: $20

If 5 heads: $25

If 6 heads: $80

If $X$ is the amount Alice pays, find $E(X)$

I have no idea how to start this

EDIT:

Okay so for 0 heads, we have $binom60 * (frac12)^0 * (frac12)^6 = frac164$

1 heads: $binom61 * (frac12)^1 * (frac12)^5 = frac332$

2 heads: $binom62 * (frac12)^2 * (frac12)^4= frac1564$

3 heads: $binom63 * (frac12)^3 * (frac12)^3 = frac516$

4 heads: $binom64 * (frac12)^4 * (frac12)^2 = frac1564$

5 heads: $binom65 * (frac12)^5 * (frac12)^1 = frac332$

6 heads: $binom66 * (frac12)^6 * (frac12)^0 = frac164$

Then I take each of these values and multiply it with X right? So for example, for 0 heads which is $frac164$, I multiply it with 200, then add it with $frac332 * 50$ for 1 heads, etc.?

probability discrete-mathematics

share|cite|improve this question

edited 2 days ago

Stuy

asked 2 days ago

StuyStuy

4351412

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Are you able to calculate the probability for no head, 1 head, 2 heads, ...? Key word: Binomial distribution. Now it´s your turn!
  $endgroup$
  – callculus
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Well the thing I'm confused about is whether or not we have bernoulli trials. It says we toss 6 coins, rather than 1 coin 6 times in a row, so we're just tossing a total of 6 coins in one go. Each coin has a 1/2 chance of being heads or tails, and since there are 6 coins, then the probability is $frac12^6$?
  $endgroup$
  – Stuy
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes, tossing the coins one after another is the same as tossing them all at the same time.
  $endgroup$
  – saulspatz
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Stuy Yes, but this is only a part of the binomial distribution. You forgot the binomial coefficient. What is the prob. that $1$ coin is head?
  $endgroup$
  – callculus
  2 days ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Oh okay, so since we do have Bernoulli Trials, for 0 heads it would be $binom60 * (frac12)^0*(frac12)^5$?
  $endgroup$
  – Stuy
  2 days ago
  

  
  
  
  

 | 
show 4 more comments

$begingroup$

Six coins are tossed. Alice pays Betty according to the following table:

If no heads: $200

If 1 head: $50

If 2 heads: $10

If 3 heads: $5

If 4 heads: $20

If 5 heads: $25

If 6 heads: $80

If $X$ is the amount Alice pays, find $E(X)$

I have no idea how to start this

EDIT:

Okay so for 0 heads, we have $binom60 * (frac12)^0 * (frac12)^6 = frac164$

1 heads: $binom61 * (frac12)^1 * (frac12)^5 = frac332$

2 heads: $binom62 * (frac12)^2 * (frac12)^4= frac1564$

3 heads: $binom63 * (frac12)^3 * (frac12)^3 = frac516$

4 heads: $binom64 * (frac12)^4 * (frac12)^2 = frac1564$

5 heads: $binom65 * (frac12)^5 * (frac12)^1 = frac332$

6 heads: $binom66 * (frac12)^6 * (frac12)^0 = frac164$

Then I take each of these values and multiply it with X right? So for example, for 0 heads which is $frac164$, I multiply it with 200, then add it with $frac332 * 50$ for 1 heads, etc.?

probability discrete-mathematics

share|cite|improve this question

edited 2 days ago

Stuy

asked 2 days ago

StuyStuy

4351412

$endgroup$

share|cite|improve this question

edited 2 days ago

Stuy

asked 2 days ago

StuyStuy

4351412

share|cite|improve this question

edited 2 days ago

Stuy

asked 2 days ago

StuyStuy

4351412

asked 2 days ago

StuyStuy

4351412

asked 2 days ago

StuyStuy

4351412