Conditional Probability Question given 3 Coins
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I had this question as a bonus problem on a previous exam and thought it was interesting, but I had no idea how to tackle it.
There are three coins in a box. One is a two-headed coin, another is a fair coin, and the third is a biased coin that comes up heads 75% of the time. When one of the three coins is selected at random and flipped, it shows heads. What is the probability that it was the first coin?
probability statistics
asked Dec 7 '18 at 0:24
HDemaHDema
12
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add a comment |
$begingroup$
I had this question as a bonus problem on a previous exam and thought it was interesting, but I had no idea how to tackle it.
There are three coins in a box. One is a two-headed coin, another is a fair coin, and the third is a biased coin that comes up heads 75% of the time. When one of the three coins is selected at random and flipped, it shows heads. What is the probability that it was the first coin?
probability statistics
asked Dec 7 '18 at 0:24
HDemaHDema
12
$endgroup$
1
$begingroup$
This is a problem for Bayes' theorem. No more no less.
$endgroup$
– Makina
Dec 7 '18 at 0:46
add a comment |
$begingroup$
I had this question as a bonus problem on a previous exam and thought it was interesting, but I had no idea how to tackle it.
There are three coins in a box. One is a two-headed coin, another is a fair coin, and the third is a biased coin that comes up heads 75% of the time. When one of the three coins is selected at random and flipped, it shows heads. What is the probability that it was the first coin?
probability statistics
asked Dec 7 '18 at 0:24
HDemaHDema
12
$endgroup$
I had this question as a bonus problem on a previous exam and thought it was interesting, but I had no idea how to tackle it.
There are three coins in a box. One is a two-headed coin, another is a fair coin, and the third is a biased coin that comes up heads 75% of the time. When one of the three coins is selected at random and flipped, it shows heads. What is the probability that it was the first coin?
probability statistics
probability statistics
asked Dec 7 '18 at 0:24
HDemaHDema
12
asked Dec 7 '18 at 0:24
HDemaHDema
12
asked Dec 7 '18 at 0:24
HDemaHDema
12
asked Dec 7 '18 at 0:24
HDemaHDema
12
asked Dec 7 '18 at 0:24
HDemaHDema
12
12
1
$begingroup$
This is a problem for Bayes' theorem. No more no less.
$endgroup$
– Makina
Dec 7 '18 at 0:46
add a comment |
1
$begingroup$
This is a problem for Bayes' theorem. No more no less.
$endgroup$
– Makina
Dec 7 '18 at 0:46
1
1
$begingroup$
This is a problem for Bayes' theorem. No more no less.
$endgroup$
– Makina
Dec 7 '18 at 0:46
$begingroup$
This is a problem for Bayes' theorem. No more no less.
$endgroup$
– Makina
Dec 7 '18 at 0:46
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
One possibility to think about it "in a very naive way" is as follows. The coins are numbered as 1, 2, 3 - say. Instead of the given experiment, we can imagine an equivalent one. Consider a hat with $12$ tickets in it, labeled as
$$
1H, 1H, 1H, 1H;
2H, 2H, 2T, 2T;
3H, 3H, 3H, 3T.
$$
Choosing a coin $k$ has the same probability as choosing a ticket $k?$ from the hat. For each fixed $k$, getting head (H) for the coin $k$ has the same probability as picking the ticket $kH$.
Now we pass to the conditional probability. We have only
$$color{red}{1H, 1H, 1H, 1H}; 2H, 2H; 3H, 3H, 3H
$$
at our disposal. Picking $1H$ has then probability $$frac 49 .$$
answered Dec 7 '18 at 1:07
dan_fuleadan_fulea
6,4781312
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add a comment |
$begingroup$
Bayes' theorem approach:
$P(A_1), P(A_2)$ and $P(A_3) = frac{1}{3}$ - choice of coins
$P(B)$ = get heads
You are asked a question to find $P(A_1 | B)$
$P(A_1 | B) = frac{P(A_1)*P(B | A_1)}{P(B)} = frac{P(A_1)*P(B | A_1)}{P(B | A_1)*P(A_1) + P(B | A_2)*P(A_2) + P(B | A_3)*P(A_3)} = frac{4}{9} $
answered Dec 7 '18 at 1:03
MakinaMakina
1,1581316
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2 Answers
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active
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2 Answers
2
active
oldest
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active
oldest
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active
oldest
votes
$begingroup$
One possibility to think about it "in a very naive way" is as follows. The coins are numbered as 1, 2, 3 - say. Instead of the given experiment, we can imagine an equivalent one. Consider a hat with $12$ tickets in it, labeled as
$$
1H, 1H, 1H, 1H;
2H, 2H, 2T, 2T;
3H, 3H, 3H, 3T.
$$
Choosing a coin $k$ has the same probability as choosing a ticket $k?$ from the hat. For each fixed $k$, getting head (H) for the coin $k$ has the same probability as picking the ticket $kH$.
Now we pass to the conditional probability. We have only
$$color{red}{1H, 1H, 1H, 1H}; 2H, 2H; 3H, 3H, 3H
$$
at our disposal. Picking $1H$ has then probability $$frac 49 .$$
answered Dec 7 '18 at 1:07
dan_fuleadan_fulea
6,4781312
$endgroup$
add a comment |
$begingroup$
One possibility to think about it "in a very naive way" is as follows. The coins are numbered as 1, 2, 3 - say. Instead of the given experiment, we can imagine an equivalent one. Consider a hat with $12$ tickets in it, labeled as
$$
1H, 1H, 1H, 1H;
2H, 2H, 2T, 2T;
3H, 3H, 3H, 3T.
$$
Choosing a coin $k$ has the same probability as choosing a ticket $k?$ from the hat. For each fixed $k$, getting head (H) for the coin $k$ has the same probability as picking the ticket $kH$.
Now we pass to the conditional probability. We have only
$$color{red}{1H, 1H, 1H, 1H}; 2H, 2H; 3H, 3H, 3H
$$
at our disposal. Picking $1H$ has then probability $$frac 49 .$$
answered Dec 7 '18 at 1:07
dan_fuleadan_fulea
6,4781312
$endgroup$
add a comment |
$begingroup$
One possibility to think about it "in a very naive way" is as follows. The coins are numbered as 1, 2, 3 - say. Instead of the given experiment, we can imagine an equivalent one. Consider a hat with $12$ tickets in it, labeled as
$$
1H, 1H, 1H, 1H;
2H, 2H, 2T, 2T;
3H, 3H, 3H, 3T.
$$
Choosing a coin $k$ has the same probability as choosing a ticket $k?$ from the hat. For each fixed $k$, getting head (H) for the coin $k$ has the same probability as picking the ticket $kH$.
Now we pass to the conditional probability. We have only
$$color{red}{1H, 1H, 1H, 1H}; 2H, 2H; 3H, 3H, 3H
$$
at our disposal. Picking $1H$ has then probability $$frac 49 .$$
answered Dec 7 '18 at 1:07
dan_fuleadan_fulea
6,4781312
$endgroup$
One possibility to think about it "in a very naive way" is as follows. The coins are numbered as 1, 2, 3 - say. Instead of the given experiment, we can imagine an equivalent one. Consider a hat with $12$ tickets in it, labeled as
$$
1H, 1H, 1H, 1H;
2H, 2H, 2T, 2T;
3H, 3H, 3H, 3T.
$$
Choosing a coin $k$ has the same probability as choosing a ticket $k?$ from the hat. For each fixed $k$, getting head (H) for the coin $k$ has the same probability as picking the ticket $kH$.
Now we pass to the conditional probability. We have only
$$color{red}{1H, 1H, 1H, 1H}; 2H, 2H; 3H, 3H, 3H
$$
at our disposal. Picking $1H$ has then probability $$frac 49 .$$
answered Dec 7 '18 at 1:07
dan_fuleadan_fulea
6,4781312
answered Dec 7 '18 at 1:07
dan_fuleadan_fulea
6,4781312
answered Dec 7 '18 at 1:07
dan_fuleadan_fulea
6,4781312
answered Dec 7 '18 at 1:07
dan_fuleadan_fulea
6,4781312
6,4781312
add a comment |
add a comment |
$begingroup$
Bayes' theorem approach:
$P(A_1), P(A_2)$ and $P(A_3) = frac{1}{3}$ - choice of coins
$P(B)$ = get heads
You are asked a question to find $P(A_1 | B)$
$P(A_1 | B) = frac{P(A_1)*P(B | A_1)}{P(B)} = frac{P(A_1)*P(B | A_1)}{P(B | A_1)*P(A_1) + P(B | A_2)*P(A_2) + P(B | A_3)*P(A_3)} = frac{4}{9} $
answered Dec 7 '18 at 1:03
MakinaMakina
1,1581316
$endgroup$
add a comment |
$begingroup$
Bayes' theorem approach:
$P(A_1), P(A_2)$ and $P(A_3) = frac{1}{3}$ - choice of coins
$P(B)$ = get heads
You are asked a question to find $P(A_1 | B)$
$P(A_1 | B) = frac{P(A_1)*P(B | A_1)}{P(B)} = frac{P(A_1)*P(B | A_1)}{P(B | A_1)*P(A_1) + P(B | A_2)*P(A_2) + P(B | A_3)*P(A_3)} = frac{4}{9} $
answered Dec 7 '18 at 1:03
MakinaMakina
1,1581316
$endgroup$
add a comment |
$begingroup$
Bayes' theorem approach:
$P(A_1), P(A_2)$ and $P(A_3) = frac{1}{3}$ - choice of coins
$P(B)$ = get heads
You are asked a question to find $P(A_1 | B)$
$P(A_1 | B) = frac{P(A_1)*P(B | A_1)}{P(B)} = frac{P(A_1)*P(B | A_1)}{P(B | A_1)*P(A_1) + P(B | A_2)*P(A_2) + P(B | A_3)*P(A_3)} = frac{4}{9} $
answered Dec 7 '18 at 1:03
MakinaMakina
1,1581316
$endgroup$
Bayes' theorem approach:
$P(A_1), P(A_2)$ and $P(A_3) = frac{1}{3}$ - choice of coins
$P(B)$ = get heads
You are asked a question to find $P(A_1 | B)$
$P(A_1 | B) = frac{P(A_1)*P(B | A_1)}{P(B)} = frac{P(A_1)*P(B | A_1)}{P(B | A_1)*P(A_1) + P(B | A_2)*P(A_2) + P(B | A_3)*P(A_3)} = frac{4}{9} $
answered Dec 7 '18 at 1:03
MakinaMakina
1,1581316
edited Dec 7 '18 at 1:08
answered Dec 7 '18 at 1:03
MakinaMakina
1,1581316
answered Dec 7 '18 at 1:03
MakinaMakina
1,1581316
answered Dec 7 '18 at 1:03
MakinaMakina
1,1581316
1,1581316
add a comment |
add a comment |
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1
$begingroup$
This is a problem for Bayes' theorem. No more no less.
$endgroup$
– Makina
Dec 7 '18 at 0:46