Tricky coin probability

The name of the pictureThe name of the pictureThe name of the pictureClash Royale CLAN TAG#URR8PPP











up vote
3
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Can someone please help, I am getting the wrong answer.



Consider four coins labelled as 1, 2, 3 and 4. Suppose that the probability of obtaining a ‘head’ in a single toss of the 𝑖-𝑡h coin is $𝑖/4,, 𝑖 = 1, 2, 3, 4.$ A coin is chosen uniformly at random and flipped.



Given that the flip resulted in a ‘head’, the conditional probability that the coin was labelled either 1 or 2 equals.



I tried doing this



P(coin 1 or 2 | tails) =
P(coin 1 or 2 And Tails)/P(tails)
=(1/4)(3/4)+(1/4)(2/4) / 1/4(3/4)+(1/4)(2/4)+(1/4)(1/4)
=5/6



In improved notation:
$$P(C_1 cup C_2 | T) = P((C_1 cup C_2)T)/P(T) \
= frac(1/4)(3/4) + (1/4)(2/4)
(1/4)(3/4) + (1/4)(2/4) + (1/4)(1/4) = 5/6.$$



But this is the wrong answer as the answer is one of 1/10,2/10,3/10,4/10.
How to do this?




probability statistics statistical-inference




share|cite|improve this question






















  • Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
    – José Carlos Santos
    Sep 1 at 17:46










  • I have tried to improve your notation. Please check that I have not changed what you mean to say. Change as necessary.
    – BruceET
    Sep 1 at 18:05







  • 4




    Hmmm I'm confused, it's given the flip resulted in a head so shouldn't you be solving for $P(C_1 cup C_2 |H)$ ?
    – HJ_beginner
    Sep 1 at 18:08






  • 1




    You solved it correctly, only as @HJ_beginner mentions, you calculated the wrong object.
    – Stan Tendijck
    Sep 1 at 18:19










  • Yes I actually read the question wrong and understood 'chosen and flipped' as the coin is tossed and then turned the other side to record heads
    – Rahul Deora
    Sep 2 at 7:54














up vote
3
down vote

favorite












Can someone please help, I am getting the wrong answer.



Consider four coins labelled as 1, 2, 3 and 4. Suppose that the probability of obtaining a ‘head’ in a single toss of the 𝑖-𝑡h coin is $𝑖/4,, 𝑖 = 1, 2, 3, 4.$ A coin is chosen uniformly at random and flipped.



Given that the flip resulted in a ‘head’, the conditional probability that the coin was labelled either 1 or 2 equals.



I tried doing this



P(coin 1 or 2 | tails) =
P(coin 1 or 2 And Tails)/P(tails)
=(1/4)(3/4)+(1/4)(2/4) / 1/4(3/4)+(1/4)(2/4)+(1/4)(1/4)
=5/6



In improved notation:
$$P(C_1 cup C_2 | T) = P((C_1 cup C_2)T)/P(T) \
= frac(1/4)(3/4) + (1/4)(2/4)
(1/4)(3/4) + (1/4)(2/4) + (1/4)(1/4) = 5/6.$$



But this is the wrong answer as the answer is one of 1/10,2/10,3/10,4/10.
How to do this?




probability statistics statistical-inference




share|cite|improve this question






















  • Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
    – José Carlos Santos
    Sep 1 at 17:46










  • I have tried to improve your notation. Please check that I have not changed what you mean to say. Change as necessary.
    – BruceET
    Sep 1 at 18:05







  • 4




    Hmmm I'm confused, it's given the flip resulted in a head so shouldn't you be solving for $P(C_1 cup C_2 |H)$ ?
    – HJ_beginner
    Sep 1 at 18:08






  • 1




    You solved it correctly, only as @HJ_beginner mentions, you calculated the wrong object.
    – Stan Tendijck
    Sep 1 at 18:19










  • Yes I actually read the question wrong and understood 'chosen and flipped' as the coin is tossed and then turned the other side to record heads
    – Rahul Deora
    Sep 2 at 7:54












up vote
3
down vote

favorite









up vote
3
down vote

favorite











Can someone please help, I am getting the wrong answer.



Consider four coins labelled as 1, 2, 3 and 4. Suppose that the probability of obtaining a ‘head’ in a single toss of the 𝑖-𝑡h coin is $𝑖/4,, 𝑖 = 1, 2, 3, 4.$ A coin is chosen uniformly at random and flipped.



Given that the flip resulted in a ‘head’, the conditional probability that the coin was labelled either 1 or 2 equals.



I tried doing this



P(coin 1 or 2 | tails) =
P(coin 1 or 2 And Tails)/P(tails)
=(1/4)(3/4)+(1/4)(2/4) / 1/4(3/4)+(1/4)(2/4)+(1/4)(1/4)
=5/6



In improved notation:
$$P(C_1 cup C_2 | T) = P((C_1 cup C_2)T)/P(T) \
= frac(1/4)(3/4) + (1/4)(2/4)
(1/4)(3/4) + (1/4)(2/4) + (1/4)(1/4) = 5/6.$$



But this is the wrong answer as the answer is one of 1/10,2/10,3/10,4/10.
How to do this?




probability statistics statistical-inference




share|cite|improve this question














Can someone please help, I am getting the wrong answer.



Consider four coins labelled as 1, 2, 3 and 4. Suppose that the probability of obtaining a ‘head’ in a single toss of the 𝑖-𝑡h coin is $𝑖/4,, 𝑖 = 1, 2, 3, 4.$ A coin is chosen uniformly at random and flipped.



Given that the flip resulted in a ‘head’, the conditional probability that the coin was labelled either 1 or 2 equals.



I tried doing this



P(coin 1 or 2 | tails) =
P(coin 1 or 2 And Tails)/P(tails)
=(1/4)(3/4)+(1/4)(2/4) / 1/4(3/4)+(1/4)(2/4)+(1/4)(1/4)
=5/6



In improved notation:
$$P(C_1 cup C_2 | T) = P((C_1 cup C_2)T)/P(T) \
= frac(1/4)(3/4) + (1/4)(2/4)
(1/4)(3/4) + (1/4)(2/4) + (1/4)(1/4) = 5/6.$$



But this is the wrong answer as the answer is one of 1/10,2/10,3/10,4/10.
How to do this?





probability statistics statistical-inference





share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Sep 1 at 18:09









BruceET

33.7k71440




33.7k71440










asked Sep 1 at 17:43









Rahul Deora

252




252











  • Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
    – José Carlos Santos
    Sep 1 at 17:46










  • I have tried to improve your notation. Please check that I have not changed what you mean to say. Change as necessary.
    – BruceET
    Sep 1 at 18:05







  • 4




    Hmmm I'm confused, it's given the flip resulted in a head so shouldn't you be solving for $P(C_1 cup C_2 |H)$ ?
    – HJ_beginner
    Sep 1 at 18:08






  • 1




    You solved it correctly, only as @HJ_beginner mentions, you calculated the wrong object.
    – Stan Tendijck
    Sep 1 at 18:19










  • Yes I actually read the question wrong and understood 'chosen and flipped' as the coin is tossed and then turned the other side to record heads
    – Rahul Deora
    Sep 2 at 7:54
















  • Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
    – José Carlos Santos
    Sep 1 at 17:46










  • I have tried to improve your notation. Please check that I have not changed what you mean to say. Change as necessary.
    – BruceET
    Sep 1 at 18:05







  • 4




    Hmmm I'm confused, it's given the flip resulted in a head so shouldn't you be solving for $P(C_1 cup C_2 |H)$ ?
    – HJ_beginner
    Sep 1 at 18:08






  • 1




    You solved it correctly, only as @HJ_beginner mentions, you calculated the wrong object.
    – Stan Tendijck
    Sep 1 at 18:19










  • Yes I actually read the question wrong and understood 'chosen and flipped' as the coin is tossed and then turned the other side to record heads
    – Rahul Deora
    Sep 2 at 7:54















Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
– José Carlos Santos
Sep 1 at 17:46




Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
– José Carlos Santos
Sep 1 at 17:46












I have tried to improve your notation. Please check that I have not changed what you mean to say. Change as necessary.
– BruceET
Sep 1 at 18:05





I have tried to improve your notation. Please check that I have not changed what you mean to say. Change as necessary.
– BruceET
Sep 1 at 18:05





4




4




Hmmm I'm confused, it's given the flip resulted in a head so shouldn't you be solving for $P(C_1 cup C_2 |H)$ ?
– HJ_beginner
Sep 1 at 18:08




Hmmm I'm confused, it's given the flip resulted in a head so shouldn't you be solving for $P(C_1 cup C_2 |H)$ ?
– HJ_beginner
Sep 1 at 18:08




1




1




You solved it correctly, only as @HJ_beginner mentions, you calculated the wrong object.
– Stan Tendijck
Sep 1 at 18:19




You solved it correctly, only as @HJ_beginner mentions, you calculated the wrong object.
– Stan Tendijck
Sep 1 at 18:19












Yes I actually read the question wrong and understood 'chosen and flipped' as the coin is tossed and then turned the other side to record heads
– Rahul Deora
Sep 2 at 7:54




Yes I actually read the question wrong and understood 'chosen and flipped' as the coin is tossed and then turned the other side to record heads
– Rahul Deora
Sep 2 at 7:54










2 Answers
2






active

oldest

votes

















up vote
2
down vote



accepted










Hint: The probability that a head is flipped is



$sum_i=1^4 P(text"coin i is chosen") cdot P(text"flip resulted in a ‘head’, ")$



$=frac14cdot frac14+frac14cdot frac24+frac14cdot frac34+frac14cdot frac44=frac58$



And the numerator is $(1/4)(colorred1/4) + (1/4)(2/4)$ since coin 1 and coin 2 are involved. You have used the probability of the third coin.



Therefore the answer of the question is $frac(1/4)(1/4) + (1/4)(2/4)frac58=frac3/165/8=frac310$






share|cite|improve this answer




















  • Thanks! I actually read the question wrong and understood 'chosen and flipped' as the coin is tossed and then turned the other side to record heads
    – Rahul Deora
    Sep 2 at 4:39

















up vote
3
down vote













First we find:$$P(H)=sum_i=1^4P(Hmid C_i)P(C_i)=frac14frac14+frac24frac14+frac34frac14+frac44frac14=frac1016=frac58tag1$$



Then:



$$P(C_1cup C_2mid H)P(H)=P((C_1cup C_2)cap H)=$$$$P(C_1cap H)+P(C_2cap H)=P(Hmid C_1)P(C_1)+P(Hmid C_2)P(C_2)=frac14frac14+frac12frac14tag2$$



Combining $(1)$ and $(2)$ we find: $$P(C_1cup C_2mid H)=frac85left(frac14frac14+frac12frac14right)=frac310$$






share|cite|improve this answer




















  • Solving the right problem (+1)
    – BruceET
    Sep 1 at 18:23










  • Thanks! I actually read the question wrong and understood 'chosen and flipped' as the coin is tossed and then turned the other side to record heads.
    – Rahul Deora
    Sep 2 at 4:39










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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
2
down vote



accepted










Hint: The probability that a head is flipped is



$sum_i=1^4 P(text"coin i is chosen") cdot P(text"flip resulted in a ‘head’, ")$



$=frac14cdot frac14+frac14cdot frac24+frac14cdot frac34+frac14cdot frac44=frac58$



And the numerator is $(1/4)(colorred1/4) + (1/4)(2/4)$ since coin 1 and coin 2 are involved. You have used the probability of the third coin.



Therefore the answer of the question is $frac(1/4)(1/4) + (1/4)(2/4)frac58=frac3/165/8=frac310$






share|cite|improve this answer




















  • Thanks! I actually read the question wrong and understood 'chosen and flipped' as the coin is tossed and then turned the other side to record heads
    – Rahul Deora
    Sep 2 at 4:39














up vote
2
down vote



accepted










Hint: The probability that a head is flipped is



$sum_i=1^4 P(text"coin i is chosen") cdot P(text"flip resulted in a ‘head’, ")$



$=frac14cdot frac14+frac14cdot frac24+frac14cdot frac34+frac14cdot frac44=frac58$



And the numerator is $(1/4)(colorred1/4) + (1/4)(2/4)$ since coin 1 and coin 2 are involved. You have used the probability of the third coin.



Therefore the answer of the question is $frac(1/4)(1/4) + (1/4)(2/4)frac58=frac3/165/8=frac310$






share|cite|improve this answer




















  • Thanks! I actually read the question wrong and understood 'chosen and flipped' as the coin is tossed and then turned the other side to record heads
    – Rahul Deora
    Sep 2 at 4:39












up vote
2
down vote



accepted







up vote
2
down vote



accepted






Hint: The probability that a head is flipped is



$sum_i=1^4 P(text"coin i is chosen") cdot P(text"flip resulted in a ‘head’, ")$



$=frac14cdot frac14+frac14cdot frac24+frac14cdot frac34+frac14cdot frac44=frac58$



And the numerator is $(1/4)(colorred1/4) + (1/4)(2/4)$ since coin 1 and coin 2 are involved. You have used the probability of the third coin.



Therefore the answer of the question is $frac(1/4)(1/4) + (1/4)(2/4)frac58=frac3/165/8=frac310$






share|cite|improve this answer












Hint: The probability that a head is flipped is



$sum_i=1^4 P(text"coin i is chosen") cdot P(text"flip resulted in a ‘head’, ")$



$=frac14cdot frac14+frac14cdot frac24+frac14cdot frac34+frac14cdot frac44=frac58$



And the numerator is $(1/4)(colorred1/4) + (1/4)(2/4)$ since coin 1 and coin 2 are involved. You have used the probability of the third coin.



Therefore the answer of the question is $frac(1/4)(1/4) + (1/4)(2/4)frac58=frac3/165/8=frac310$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Sep 1 at 18:19









callculus

16.7k31427




16.7k31427











  • Thanks! I actually read the question wrong and understood 'chosen and flipped' as the coin is tossed and then turned the other side to record heads
    – Rahul Deora
    Sep 2 at 4:39
















  • Thanks! I actually read the question wrong and understood 'chosen and flipped' as the coin is tossed and then turned the other side to record heads
    – Rahul Deora
    Sep 2 at 4:39















Thanks! I actually read the question wrong and understood 'chosen and flipped' as the coin is tossed and then turned the other side to record heads
– Rahul Deora
Sep 2 at 4:39




Thanks! I actually read the question wrong and understood 'chosen and flipped' as the coin is tossed and then turned the other side to record heads
– Rahul Deora
Sep 2 at 4:39










up vote
3
down vote













First we find:$$P(H)=sum_i=1^4P(Hmid C_i)P(C_i)=frac14frac14+frac24frac14+frac34frac14+frac44frac14=frac1016=frac58tag1$$



Then:



$$P(C_1cup C_2mid H)P(H)=P((C_1cup C_2)cap H)=$$$$P(C_1cap H)+P(C_2cap H)=P(Hmid C_1)P(C_1)+P(Hmid C_2)P(C_2)=frac14frac14+frac12frac14tag2$$



Combining $(1)$ and $(2)$ we find: $$P(C_1cup C_2mid H)=frac85left(frac14frac14+frac12frac14right)=frac310$$






share|cite|improve this answer




















  • Solving the right problem (+1)
    – BruceET
    Sep 1 at 18:23










  • Thanks! I actually read the question wrong and understood 'chosen and flipped' as the coin is tossed and then turned the other side to record heads.
    – Rahul Deora
    Sep 2 at 4:39














up vote
3
down vote













First we find:$$P(H)=sum_i=1^4P(Hmid C_i)P(C_i)=frac14frac14+frac24frac14+frac34frac14+frac44frac14=frac1016=frac58tag1$$



Then:



$$P(C_1cup C_2mid H)P(H)=P((C_1cup C_2)cap H)=$$$$P(C_1cap H)+P(C_2cap H)=P(Hmid C_1)P(C_1)+P(Hmid C_2)P(C_2)=frac14frac14+frac12frac14tag2$$



Combining $(1)$ and $(2)$ we find: $$P(C_1cup C_2mid H)=frac85left(frac14frac14+frac12frac14right)=frac310$$






share|cite|improve this answer




















  • Solving the right problem (+1)
    – BruceET
    Sep 1 at 18:23










  • Thanks! I actually read the question wrong and understood 'chosen and flipped' as the coin is tossed and then turned the other side to record heads.
    – Rahul Deora
    Sep 2 at 4:39












up vote
3
down vote










up vote
3
down vote









First we find:$$P(H)=sum_i=1^4P(Hmid C_i)P(C_i)=frac14frac14+frac24frac14+frac34frac14+frac44frac14=frac1016=frac58tag1$$



Then:



$$P(C_1cup C_2mid H)P(H)=P((C_1cup C_2)cap H)=$$$$P(C_1cap H)+P(C_2cap H)=P(Hmid C_1)P(C_1)+P(Hmid C_2)P(C_2)=frac14frac14+frac12frac14tag2$$



Combining $(1)$ and $(2)$ we find: $$P(C_1cup C_2mid H)=frac85left(frac14frac14+frac12frac14right)=frac310$$






share|cite|improve this answer












First we find:$$P(H)=sum_i=1^4P(Hmid C_i)P(C_i)=frac14frac14+frac24frac14+frac34frac14+frac44frac14=frac1016=frac58tag1$$



Then:



$$P(C_1cup C_2mid H)P(H)=P((C_1cup C_2)cap H)=$$$$P(C_1cap H)+P(C_2cap H)=P(Hmid C_1)P(C_1)+P(Hmid C_2)P(C_2)=frac14frac14+frac12frac14tag2$$



Combining $(1)$ and $(2)$ we find: $$P(C_1cup C_2mid H)=frac85left(frac14frac14+frac12frac14right)=frac310$$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Sep 1 at 18:15









drhab

88.7k541120




88.7k541120











  • Solving the right problem (+1)
    – BruceET
    Sep 1 at 18:23










  • Thanks! I actually read the question wrong and understood 'chosen and flipped' as the coin is tossed and then turned the other side to record heads.
    – Rahul Deora
    Sep 2 at 4:39
















  • Solving the right problem (+1)
    – BruceET
    Sep 1 at 18:23










  • Thanks! I actually read the question wrong and understood 'chosen and flipped' as the coin is tossed and then turned the other side to record heads.
    – Rahul Deora
    Sep 2 at 4:39















Solving the right problem (+1)
– BruceET
Sep 1 at 18:23




Solving the right problem (+1)
– BruceET
Sep 1 at 18:23












Thanks! I actually read the question wrong and understood 'chosen and flipped' as the coin is tossed and then turned the other side to record heads.
– Rahul Deora
Sep 2 at 4:39




Thanks! I actually read the question wrong and understood 'chosen and flipped' as the coin is tossed and then turned the other side to record heads.
– Rahul Deora
Sep 2 at 4:39

















 

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