probability of choosing a biased coin C which has probability 3/15 of getting heads assuming we got head on...
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:
$$1/15 * 1/2 = 1/30$$
probability discrete-mathematics conditional-probability
asked 1 hour ago
hussain sagar
776
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:
$$1/15 * 1/2 = 1/30$$
probability discrete-mathematics conditional-probability
asked 1 hour ago
hussain sagar
776
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:
$$1/15 * 1/2 = 1/30$$
probability discrete-mathematics conditional-probability
asked 1 hour ago
hussain sagar
776
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:
$$1/15 * 1/2 = 1/30$$
probability discrete-mathematics conditional-probability
probability discrete-mathematics conditional-probability
asked 1 hour ago
hussain sagar
776
asked 1 hour ago
hussain sagar
776
asked 1 hour ago
hussain sagar
776
asked 1 hour ago
hussain sagar
776
asked 1 hour ago
hussain sagar
776
776
add a comment |
add a comment |
2 Answers
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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
$$
frac{1/30}{1/12 + 1/20 + 1/30}
= frac{1}{5/2+3/2 + 1}
= frac15.
$$
answered 1 hour ago
gt6989b
32.9k22452
Thank you, this makes a lot of sense.
– hussain sagar
1 hour ago
@hussainsagar you are welcome
– gt6989b
1 hour ago
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
$$P(C|H)= frac{P(H|C)P(C)}{P(H)}=frac{P(H|C)P(C)}{P(Hcap A)+P(H cap B)+P(H cap C)}$$
answered 1 hour ago
Siong Thye Goh
98.4k1463116
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
$$
frac{1/30}{1/12 + 1/20 + 1/30}
= frac{1}{5/2+3/2 + 1}
= frac15.
$$
answered 1 hour ago
gt6989b
32.9k22452
Thank you, this makes a lot of sense.
– hussain sagar
1 hour ago
@hussainsagar you are welcome
– gt6989b
1 hour ago
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
$$
frac{1/30}{1/12 + 1/20 + 1/30}
= frac{1}{5/2+3/2 + 1}
= frac15.
$$
answered 1 hour ago
gt6989b
32.9k22452
Thank you, this makes a lot of sense.
– hussain sagar
1 hour ago
@hussainsagar you are welcome
– gt6989b
1 hour ago
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
$$
frac{1/30}{1/12 + 1/20 + 1/30}
= frac{1}{5/2+3/2 + 1}
= frac15.
$$
answered 1 hour ago
gt6989b
32.9k22452
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
$$
frac{1/30}{1/12 + 1/20 + 1/30}
= frac{1}{5/2+3/2 + 1}
= frac15.
$$
answered 1 hour ago
gt6989b
32.9k22452
answered 1 hour ago
gt6989b
32.9k22452
answered 1 hour ago
gt6989b
32.9k22452
answered 1 hour ago
gt6989b
32.9k22452
32.9k22452
Thank you, this makes a lot of sense.
– hussain sagar
1 hour ago
@hussainsagar you are welcome
– gt6989b
1 hour ago
add a comment |
Thank you, this makes a lot of sense.
– hussain sagar
1 hour ago
@hussainsagar you are welcome
– gt6989b
1 hour ago
Thank you, this makes a lot of sense.
– hussain sagar
1 hour ago
Thank you, this makes a lot of sense.
– hussain sagar
1 hour ago
@hussainsagar you are welcome
– gt6989b
1 hour ago
@hussainsagar you are welcome
– gt6989b
1 hour ago
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
$$P(C|H)= frac{P(H|C)P(C)}{P(H)}=frac{P(H|C)P(C)}{P(Hcap A)+P(H cap B)+P(H cap C)}$$
answered 1 hour ago
Siong Thye Goh
98.4k1463116
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
$$P(C|H)= frac{P(H|C)P(C)}{P(H)}=frac{P(H|C)P(C)}{P(Hcap A)+P(H cap B)+P(H cap C)}$$
answered 1 hour ago
Siong Thye Goh
98.4k1463116
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
$$P(C|H)= frac{P(H|C)P(C)}{P(H)}=frac{P(H|C)P(C)}{P(Hcap A)+P(H cap B)+P(H cap C)}$$
answered 1 hour ago
Siong Thye Goh
98.4k1463116
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
$$P(C|H)= frac{P(H|C)P(C)}{P(H)}=frac{P(H|C)P(C)}{P(Hcap A)+P(H cap B)+P(H cap C)}$$
answered 1 hour ago
Siong Thye Goh
98.4k1463116
answered 1 hour ago
Siong Thye Goh
98.4k1463116
answered 1 hour ago
Siong Thye Goh
98.4k1463116
answered 1 hour ago
Siong Thye Goh
98.4k1463116
98.4k1463116
add a comment |
add a comment |
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