Probability of choosing a biased coin $C$ which has probability $3/15$ of getting heads, assuming we got head on the first toss
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Full question: there are 3 biased coins $A$, $B$, and $C$ each with probability $5/15, 3/15, 1/15$ of getting heads respectively. Also, they have probability $1/4$ for $A$, $1/4$ for $B$, and $1/2$ for $C$ of getting picked. If a coin was picked and tossed and the result was heads, what is the probability that the coin was coin $C$?
My approach: Since the coin picked was $C$ and the result was head we merely multiply the probability of both those things happening concerning $C$:
$$frac115 cdot frac12 = frac130$$
probability discrete-mathematics conditional-probability
edited Dec 24 '18 at 7:49
Asaf Karagila♦
302k32426756
asked Dec 24 '18 at 4:50
hussain sagar
806
add a comment |
Full question: there are 3 biased coins $A$, $B$, and $C$ each with probability $5/15, 3/15, 1/15$ of getting heads respectively. Also, they have probability $1/4$ for $A$, $1/4$ for $B$, and $1/2$ for $C$ of getting picked. If a coin was picked and tossed and the result was heads, what is the probability that the coin was coin $C$?
My approach: Since the coin picked was $C$ and the result was head we merely multiply the probability of both those things happening concerning $C$:
$$frac115 cdot frac12 = frac130$$
probability discrete-mathematics conditional-probability
edited Dec 24 '18 at 7:49
Asaf Karagila♦
302k32426756
asked Dec 24 '18 at 4:50
hussain sagar
806
add a comment |
Full question: there are 3 biased coins $A$, $B$, and $C$ each with probability $5/15, 3/15, 1/15$ of getting heads respectively. Also, they have probability $1/4$ for $A$, $1/4$ for $B$, and $1/2$ for $C$ of getting picked. If a coin was picked and tossed and the result was heads, what is the probability that the coin was coin $C$?
My approach: Since the coin picked was $C$ and the result was head we merely multiply the probability of both those things happening concerning $C$:
$$frac115 cdot frac12 = frac130$$
probability discrete-mathematics conditional-probability
edited Dec 24 '18 at 7:49
Asaf Karagila♦
302k32426756
asked Dec 24 '18 at 4:50
hussain sagar
806
Full question: there are 3 biased coins $A$, $B$, and $C$ each with probability $5/15, 3/15, 1/15$ of getting heads respectively. Also, they have probability $1/4$ for $A$, $1/4$ for $B$, and $1/2$ for $C$ of getting picked. If a coin was picked and tossed and the result was heads, what is the probability that the coin was coin $C$?
My approach: Since the coin picked was $C$ and the result was head we merely multiply the probability of both those things happening concerning $C$:
$$frac115 cdot frac12 = frac130$$
probability discrete-mathematics conditional-probability
probability discrete-mathematics conditional-probability
edited Dec 24 '18 at 7:49
Asaf Karagila♦
302k32426756
asked Dec 24 '18 at 4:50
hussain sagar
806
edited Dec 24 '18 at 7:49
Asaf Karagila♦
302k32426756
asked Dec 24 '18 at 4:50
hussain sagar
806
edited Dec 24 '18 at 7:49
Asaf Karagila♦
302k32426756
edited Dec 24 '18 at 7:49
Asaf Karagila♦
302k32426756
edited Dec 24 '18 at 7:49
Asaf Karagila♦
302k32426756
302k32426756
asked Dec 24 '18 at 4:50
hussain sagar
806
asked Dec 24 '18 at 4:50
hussain sagar
806
asked Dec 24 '18 at 4:50
hussain sagar
806
806
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
An intuitive argument based on Bayes' Theorem, says that getting heads was possible in one of three different ways:
- Draw $A$ with probability $1/4$ and flip heads with probability $5/15$, total chance of $1/4 times 5/15 = 1/12$.
- Draw $B$: $1/4 times 3/15 = 1/20$.
- Draw $C$: $1/2 times 1/15 = 1/30$.
Thus, the chance that $C$ was drawn is
$$
frac1/301/12 + 1/20 + 1/30
= frac15/2+3/2 + 1
= frac15.
$$
answered Dec 24 '18 at 4:55
gt6989b
33.1k22452
Thank you, this makes a lot of sense.
– hussain sagar
Dec 24 '18 at 4:57
@hussainsagar you are welcome
– gt6989b
Dec 24 '18 at 4:58
add a comment |
What you computed is the probability that coin $C$ is chosen and you get a head $H$, that is $P(H cap C)$. It is not equalty to $P(C|H)$.
Guide:
Use Bayes rule, that is
$$P(C|H)= fracC)P(C)P(H)=fracC)P(C)P(Hcap A)+P(H cap B)+P(H cap C)$$
answered Dec 24 '18 at 4:55
Siong Thye Goh
99.5k1464117
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
An intuitive argument based on Bayes' Theorem, says that getting heads was possible in one of three different ways:
- Draw $A$ with probability $1/4$ and flip heads with probability $5/15$, total chance of $1/4 times 5/15 = 1/12$.
- Draw $B$: $1/4 times 3/15 = 1/20$.
- Draw $C$: $1/2 times 1/15 = 1/30$.
Thus, the chance that $C$ was drawn is
$$
frac1/301/12 + 1/20 + 1/30
= frac15/2+3/2 + 1
= frac15.
$$
answered Dec 24 '18 at 4:55
gt6989b
33.1k22452
Thank you, this makes a lot of sense.
– hussain sagar
Dec 24 '18 at 4:57
@hussainsagar you are welcome
– gt6989b
Dec 24 '18 at 4:58
add a comment |
An intuitive argument based on Bayes' Theorem, says that getting heads was possible in one of three different ways:
- Draw $A$ with probability $1/4$ and flip heads with probability $5/15$, total chance of $1/4 times 5/15 = 1/12$.
- Draw $B$: $1/4 times 3/15 = 1/20$.
- Draw $C$: $1/2 times 1/15 = 1/30$.
Thus, the chance that $C$ was drawn is
$$
frac1/301/12 + 1/20 + 1/30
= frac15/2+3/2 + 1
= frac15.
$$
answered Dec 24 '18 at 4:55
gt6989b
33.1k22452
Thank you, this makes a lot of sense.
– hussain sagar
Dec 24 '18 at 4:57
@hussainsagar you are welcome
– gt6989b
Dec 24 '18 at 4:58
add a comment |
An intuitive argument based on Bayes' Theorem, says that getting heads was possible in one of three different ways:
- Draw $A$ with probability $1/4$ and flip heads with probability $5/15$, total chance of $1/4 times 5/15 = 1/12$.
- Draw $B$: $1/4 times 3/15 = 1/20$.
- Draw $C$: $1/2 times 1/15 = 1/30$.
Thus, the chance that $C$ was drawn is
$$
frac1/301/12 + 1/20 + 1/30
= frac15/2+3/2 + 1
= frac15.
$$
answered Dec 24 '18 at 4:55
gt6989b
33.1k22452
An intuitive argument based on Bayes' Theorem, says that getting heads was possible in one of three different ways:
- Draw $A$ with probability $1/4$ and flip heads with probability $5/15$, total chance of $1/4 times 5/15 = 1/12$.
- Draw $B$: $1/4 times 3/15 = 1/20$.
- Draw $C$: $1/2 times 1/15 = 1/30$.
Thus, the chance that $C$ was drawn is
$$
frac1/301/12 + 1/20 + 1/30
= frac15/2+3/2 + 1
= frac15.
$$
answered Dec 24 '18 at 4:55
gt6989b
33.1k22452
answered Dec 24 '18 at 4:55
gt6989b
33.1k22452
answered Dec 24 '18 at 4:55
gt6989b
33.1k22452
answered Dec 24 '18 at 4:55
gt6989b
33.1k22452
33.1k22452
Thank you, this makes a lot of sense.
– hussain sagar
Dec 24 '18 at 4:57
@hussainsagar you are welcome
– gt6989b
Dec 24 '18 at 4:58
add a comment |
Thank you, this makes a lot of sense.
– hussain sagar
Dec 24 '18 at 4:57
@hussainsagar you are welcome
– gt6989b
Dec 24 '18 at 4:58
Thank you, this makes a lot of sense.
– hussain sagar
Dec 24 '18 at 4:57
Thank you, this makes a lot of sense.
– hussain sagar
Dec 24 '18 at 4:57
@hussainsagar you are welcome
– gt6989b
Dec 24 '18 at 4:58
@hussainsagar you are welcome
– gt6989b
Dec 24 '18 at 4:58
add a comment |
What you computed is the probability that coin $C$ is chosen and you get a head $H$, that is $P(H cap C)$. It is not equalty to $P(C|H)$.
Guide:
Use Bayes rule, that is
$$P(C|H)= fracC)P(C)P(H)=fracC)P(C)P(Hcap A)+P(H cap B)+P(H cap C)$$
answered Dec 24 '18 at 4:55
Siong Thye Goh
99.5k1464117
add a comment |
What you computed is the probability that coin $C$ is chosen and you get a head $H$, that is $P(H cap C)$. It is not equalty to $P(C|H)$.
Guide:
Use Bayes rule, that is
$$P(C|H)= fracC)P(C)P(H)=fracC)P(C)P(Hcap A)+P(H cap B)+P(H cap C)$$
answered Dec 24 '18 at 4:55
Siong Thye Goh
99.5k1464117
add a comment |
What you computed is the probability that coin $C$ is chosen and you get a head $H$, that is $P(H cap C)$. It is not equalty to $P(C|H)$.
Guide:
Use Bayes rule, that is
$$P(C|H)= fracC)P(C)P(H)=fracC)P(C)P(Hcap A)+P(H cap B)+P(H cap C)$$
answered Dec 24 '18 at 4:55
Siong Thye Goh
99.5k1464117
What you computed is the probability that coin $C$ is chosen and you get a head $H$, that is $P(H cap C)$. It is not equalty to $P(C|H)$.
Guide:
Use Bayes rule, that is
$$P(C|H)= fracC)P(C)P(H)=fracC)P(C)P(Hcap A)+P(H cap B)+P(H cap C)$$
answered Dec 24 '18 at 4:55
Siong Thye Goh
99.5k1464117
answered Dec 24 '18 at 4:55
Siong Thye Goh
99.5k1464117
answered Dec 24 '18 at 4:55
Siong Thye Goh
99.5k1464117
answered Dec 24 '18 at 4:55
Siong Thye Goh
99.5k1464117
99.5k1464117
add a comment |
add a comment |
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