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A coin is thrown until the tail is above.

Question regarding solution to the problem of computing the expected number of coin flips until getting 5 in a row.The probability that tail occurs an even number of times after $491$ coin flipsWhen the probability model of an experiment is correct?Question Using law of total probabilityTwo persons $A$ & $B$ toss a coin 50 times each together. Find the probability that both of them get 'tails' at the same time.Expectation of the number of times a coin is thrown until the appearance of a second “tail”Throw a fair die repeatedly until I have thrown three sixesNot sure about the answer in elementary probability problemFlipping the coin until either head or tail has occurred 10 timesProbability of at most 1 tail? Loaded Coin Probability

2

$begingroup$

A coin is thrown until the tail is above. Which is the probability of making at most 3 throws.

The answer is :$frac{7}{8}$.

I don't understand why. What are the possible cases and what are the favorable ones?

probability

share|cite|improve this question

asked 1 hour ago

Ica SanduIca Sandu

1499

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  It may be useful to know that this is a geometrically distributed random variable.
  $endgroup$
  – Peter Foreman
  1 hour ago
  

  
  
  
  

add a comment | 

2

$begingroup$

A coin is thrown until the tail is above. Which is the probability of making at most 3 throws.

The answer is :$frac{7}{8}$.

I don't understand why. What are the possible cases and what are the favorable ones?

probability

share|cite|improve this question

asked 1 hour ago

Ica SanduIca Sandu

1499

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  It may be useful to know that this is a geometrically distributed random variable.
  $endgroup$
  – Peter Foreman
  1 hour ago
  

  
  
  
  

add a comment | 

$begingroup$

A coin is thrown until the tail is above. Which is the probability of making at most 3 throws.

The answer is :$frac{7}{8}$.

I don't understand why. What are the possible cases and what are the favorable ones?

probability

share|cite|improve this question

asked 1 hour ago

Ica SanduIca Sandu

1499

$endgroup$

share|cite|improve this question

asked 1 hour ago

Ica SanduIca Sandu

1499

share|cite|improve this question

asked 1 hour ago

Ica SanduIca Sandu

1499

asked 1 hour ago

Ica SanduIca Sandu

1499

asked 1 hour ago

Ica SanduIca Sandu

1499

3 Answers
3

active

oldest

votes

3

$begingroup$

There are three possibilities: T, HT, HHT with probabilities $1/2$, $1/2cdot1/2$, and $1/2cdot1/2cdot1/2$, resp. They sum up to $7/8$.

share|cite|improve this answer

answered 1 hour ago

Michael HoppeMichael Hoppe

11.3k31837

$endgroup$

add a comment | 

3

$begingroup$

You will need more than three throws only when head comes out thrice.

Hence

$$p=1-frac1{2^3}.$$

More generally,

$$p=1-frac1{2^n}$$ for the probability of making at most $n$ throws.

share|cite|improve this answer

edited 1 hour ago

answered 1 hour ago

Yves DaoustYves Daoust

134k676232

$endgroup$

add a comment | 

1

$begingroup$

Let $T$ be the event that a tail is obtained within 3 flips. We want to find $pr(T)$. It is a bit difficult to compute this because we have to consider obtaining tails within 1 flip, 2 flips, AND 3 flips, to which there is overlap. It is a bit easier to compute $pr(T')$ or the probability $T$ does not happen.

For event $T'$ to occur, we would need to not obtain tails within 3 flips, i.e. get heads three times in a row. Since the probability of throwing heads is $frac{1}{2}$, getting three heads in a row is $frac{1}{2^3}=frac{1}{8}$. Hence, $pr(T')=frac{1}{8}$. Since any given event either happens or doesn't, we have in general for event $A$, $$pr(A')=1-pr(A).$$ Therefore,
$$frac{1}{8}=pr(T')=1-pr(T)implies pr(T)=1-frac{1}{8}=frac{7}{8}$$.

share|cite|improve this answer

answered 1 hour ago

Jane CooperJane Cooper

536

$endgroup$

add a comment | 

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3

$begingroup$

There are three possibilities: T, HT, HHT with probabilities $1/2$, $1/2cdot1/2$, and $1/2cdot1/2cdot1/2$, resp. They sum up to $7/8$.

share|cite|improve this answer

answered 1 hour ago

Michael HoppeMichael Hoppe

11.3k31837

$endgroup$

add a comment | 

3

$begingroup$

There are three possibilities: T, HT, HHT with probabilities $1/2$, $1/2cdot1/2$, and $1/2cdot1/2cdot1/2$, resp. They sum up to $7/8$.

share|cite|improve this answer

answered 1 hour ago

Michael HoppeMichael Hoppe

11.3k31837

$endgroup$

add a comment | 

$begingroup$

There are three possibilities: T, HT, HHT with probabilities $1/2$, $1/2cdot1/2$, and $1/2cdot1/2cdot1/2$, resp. They sum up to $7/8$.

share|cite|improve this answer

answered 1 hour ago

Michael HoppeMichael Hoppe

11.3k31837

$endgroup$

share|cite|improve this answer

answered 1 hour ago

Michael HoppeMichael Hoppe

11.3k31837

answered 1 hour ago

Michael HoppeMichael Hoppe

11.3k31837

answered 1 hour ago

Michael HoppeMichael Hoppe

11.3k31837

3

$begingroup$

You will need more than three throws only when head comes out thrice.

Hence

$$p=1-frac1{2^3}.$$

More generally,

$$p=1-frac1{2^n}$$ for the probability of making at most $n$ throws.

share|cite|improve this answer

edited 1 hour ago

answered 1 hour ago

Yves DaoustYves Daoust

134k676232

$endgroup$

add a comment | 

3

$begingroup$

You will need more than three throws only when head comes out thrice.

Hence

$$p=1-frac1{2^3}.$$

More generally,

$$p=1-frac1{2^n}$$ for the probability of making at most $n$ throws.

share|cite|improve this answer

edited 1 hour ago

answered 1 hour ago

Yves DaoustYves Daoust

134k676232

$endgroup$

add a comment | 

$begingroup$

You will need more than three throws only when head comes out thrice.

Hence

$$p=1-frac1{2^3}.$$

More generally,

$$p=1-frac1{2^n}$$ for the probability of making at most $n$ throws.

share|cite|improve this answer

edited 1 hour ago

answered 1 hour ago

Yves DaoustYves Daoust

134k676232

$endgroup$

share|cite|improve this answer

edited 1 hour ago

answered 1 hour ago

Yves DaoustYves Daoust

134k676232

answered 1 hour ago

Yves DaoustYves Daoust

134k676232

answered 1 hour ago

Yves DaoustYves Daoust

134k676232

1

$begingroup$

Let $T$ be the event that a tail is obtained within 3 flips. We want to find $pr(T)$. It is a bit difficult to compute this because we have to consider obtaining tails within 1 flip, 2 flips, AND 3 flips, to which there is overlap. It is a bit easier to compute $pr(T')$ or the probability $T$ does not happen.

For event $T'$ to occur, we would need to not obtain tails within 3 flips, i.e. get heads three times in a row. Since the probability of throwing heads is $frac{1}{2}$, getting three heads in a row is $frac{1}{2^3}=frac{1}{8}$. Hence, $pr(T')=frac{1}{8}$. Since any given event either happens or doesn't, we have in general for event $A$, $$pr(A')=1-pr(A).$$ Therefore,
$$frac{1}{8}=pr(T')=1-pr(T)implies pr(T)=1-frac{1}{8}=frac{7}{8}$$.

share|cite|improve this answer

answered 1 hour ago

Jane CooperJane Cooper

536

$endgroup$

add a comment | 

1

$begingroup$

Let $T$ be the event that a tail is obtained within 3 flips. We want to find $pr(T)$. It is a bit difficult to compute this because we have to consider obtaining tails within 1 flip, 2 flips, AND 3 flips, to which there is overlap. It is a bit easier to compute $pr(T')$ or the probability $T$ does not happen.

For event $T'$ to occur, we would need to not obtain tails within 3 flips, i.e. get heads three times in a row. Since the probability of throwing heads is $frac{1}{2}$, getting three heads in a row is $frac{1}{2^3}=frac{1}{8}$. Hence, $pr(T')=frac{1}{8}$. Since any given event either happens or doesn't, we have in general for event $A$, $$pr(A')=1-pr(A).$$ Therefore,
$$frac{1}{8}=pr(T')=1-pr(T)implies pr(T)=1-frac{1}{8}=frac{7}{8}$$.

share|cite|improve this answer

answered 1 hour ago

Jane CooperJane Cooper

536

$endgroup$

add a comment | 

$begingroup$

Let $T$ be the event that a tail is obtained within 3 flips. We want to find $pr(T)$. It is a bit difficult to compute this because we have to consider obtaining tails within 1 flip, 2 flips, AND 3 flips, to which there is overlap. It is a bit easier to compute $pr(T')$ or the probability $T$ does not happen.

For event $T'$ to occur, we would need to not obtain tails within 3 flips, i.e. get heads three times in a row. Since the probability of throwing heads is $frac{1}{2}$, getting three heads in a row is $frac{1}{2^3}=frac{1}{8}$. Hence, $pr(T')=frac{1}{8}$. Since any given event either happens or doesn't, we have in general for event $A$, $$pr(A')=1-pr(A).$$ Therefore,
$$frac{1}{8}=pr(T')=1-pr(T)implies pr(T)=1-frac{1}{8}=frac{7}{8}$$.

share|cite|improve this answer

answered 1 hour ago

Jane CooperJane Cooper

536

$endgroup$

share|cite|improve this answer

answered 1 hour ago

Jane CooperJane Cooper

536

answered 1 hour ago

Jane CooperJane Cooper

536

answered 1 hour ago

Jane CooperJane Cooper

536