Suppose a channel where 0 or 1 is sent, what is probability 1 was sent given 1 was received?
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I have the following probability.
probability a $0$ is sent is $0.4$
probability a $1$ is sent is $0.6$
probability that due to noise, a $0$ is changed to a $1$ during transmission is $0.2$
probability that due to noise, a $1$ is changed to a $0$ during transmission is $0.1$
Here is what I think should be the solution:
Let $A$ be $1$ was received, $B$ be $1$ was sent
$P(B) = 0.6$
$P(A|B) = 0.6 * 0.9 = 0.54$
$P(A) = 0.6 * 0.9 + 0.4 * 0.2 = 0.62$
Thus using $$P(B|A) = frac{P(A|B)P(B)}{P(A)}$$
$P(B|A) = 52.26%$
Is my solution correct?
probability proof-verification conditional-probability
edited Dec 3 at 4:48
platty
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asked Dec 3 at 4:44
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up vote
0
down vote
favorite
I have the following probability.
probability a $0$ is sent is $0.4$
probability a $1$ is sent is $0.6$
probability that due to noise, a $0$ is changed to a $1$ during transmission is $0.2$
probability that due to noise, a $1$ is changed to a $0$ during transmission is $0.1$
Here is what I think should be the solution:
Let $A$ be $1$ was received, $B$ be $1$ was sent
$P(B) = 0.6$
$P(A|B) = 0.6 * 0.9 = 0.54$
$P(A) = 0.6 * 0.9 + 0.4 * 0.2 = 0.62$
Thus using $$P(B|A) = frac{P(A|B)P(B)}{P(A)}$$
$P(B|A) = 52.26%$
Is my solution correct?
probability proof-verification conditional-probability
edited Dec 3 at 4:48
platty
2,940318
asked Dec 3 at 4:44
Cup
276
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I have the following probability.
probability a $0$ is sent is $0.4$
probability a $1$ is sent is $0.6$
probability that due to noise, a $0$ is changed to a $1$ during transmission is $0.2$
probability that due to noise, a $1$ is changed to a $0$ during transmission is $0.1$
Here is what I think should be the solution:
Let $A$ be $1$ was received, $B$ be $1$ was sent
$P(B) = 0.6$
$P(A|B) = 0.6 * 0.9 = 0.54$
$P(A) = 0.6 * 0.9 + 0.4 * 0.2 = 0.62$
Thus using $$P(B|A) = frac{P(A|B)P(B)}{P(A)}$$
$P(B|A) = 52.26%$
Is my solution correct?
probability proof-verification conditional-probability
edited Dec 3 at 4:48
platty
2,940318
asked Dec 3 at 4:44
Cup
276
I have the following probability.
probability a $0$ is sent is $0.4$
probability a $1$ is sent is $0.6$
probability that due to noise, a $0$ is changed to a $1$ during transmission is $0.2$
probability that due to noise, a $1$ is changed to a $0$ during transmission is $0.1$
Here is what I think should be the solution:
Let $A$ be $1$ was received, $B$ be $1$ was sent
$P(B) = 0.6$
$P(A|B) = 0.6 * 0.9 = 0.54$
$P(A) = 0.6 * 0.9 + 0.4 * 0.2 = 0.62$
Thus using $$P(B|A) = frac{P(A|B)P(B)}{P(A)}$$
$P(B|A) = 52.26%$
Is my solution correct?
probability proof-verification conditional-probability
probability proof-verification conditional-probability
edited Dec 3 at 4:48
platty
2,940318
asked Dec 3 at 4:44
Cup
276
edited Dec 3 at 4:48
platty
2,940318
asked Dec 3 at 4:44
Cup
276
edited Dec 3 at 4:48
platty
2,940318
edited Dec 3 at 4:48
platty
2,940318
edited Dec 3 at 4:48
platty
2,940318
2,940318
asked Dec 3 at 4:44
Cup
276
asked Dec 3 at 4:44
Cup
276
asked Dec 3 at 4:44
Cup
276
276
add a comment |
add a comment |
1 Answer
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Not quite. $P(A | B) = 0.9$; what you wrote above is actually $P(A cap B)$. Everything else looks fine, so you should end up with $P(B mid A) = frac{0.9(0.6)}{0.62} approx 0.87$.
answered Dec 3 at 4:47
platty
2,940318
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
Not quite. $P(A | B) = 0.9$; what you wrote above is actually $P(A cap B)$. Everything else looks fine, so you should end up with $P(B mid A) = frac{0.9(0.6)}{0.62} approx 0.87$.
answered Dec 3 at 4:47
platty
2,940318
add a comment |
up vote
0
down vote
Not quite. $P(A | B) = 0.9$; what you wrote above is actually $P(A cap B)$. Everything else looks fine, so you should end up with $P(B mid A) = frac{0.9(0.6)}{0.62} approx 0.87$.
answered Dec 3 at 4:47
platty
2,940318
add a comment |
up vote
0
down vote
up vote
0
down vote
Not quite. $P(A | B) = 0.9$; what you wrote above is actually $P(A cap B)$. Everything else looks fine, so you should end up with $P(B mid A) = frac{0.9(0.6)}{0.62} approx 0.87$.
answered Dec 3 at 4:47
platty
2,940318
Not quite. $P(A | B) = 0.9$; what you wrote above is actually $P(A cap B)$. Everything else looks fine, so you should end up with $P(B mid A) = frac{0.9(0.6)}{0.62} approx 0.87$.
answered Dec 3 at 4:47
platty
2,940318
answered Dec 3 at 4:47
platty
2,940318
answered Dec 3 at 4:47
platty
2,940318
answered Dec 3 at 4:47
platty
2,940318
2,940318
add a comment |
add a comment |
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