Mixing
Conditional probability, probability of the message being authentic:
$begingroup$
As communication security becomes more and more a problem, when a message is
received, it must authenticated, by using a secret enciphering key. Sometimes, though,
it can fall into wrong hands, thus allowing an unauthentic message to appear to be
authentic. Assume that 80% of all messages received are authentic. Furthermore,
assume that only 5% of all unauthentic messages are sent using the correct key and
that all authentic messages are sent using the correct key. Find the probability that
a message is authentic given that the correct key is used.
How do I compute this event, I've been scratching my head for the last half an hour and tried to look at the problem from different angles, I tried thinking of it as $ P(A/K) $ $A$ - message is authentic and $K$ - key is correct and then tried to compute with Bayes but it makes no sense since $P(K)$ is always going to be 1 since it is specified that the identified key is authentic and $P(K/A)$ would also result in 1, since the message is authentic the key will always be correct. Am I even making sense? I feel like I'm missing something right under my nose because at first glance it doesn't seem to be such a difficult problem, so I would appreciate some help...
probability conditional-probability
asked Feb 2 at 2:49
The VirtuosoThe Virtuoso
448
$endgroup$
add a comment |
$begingroup$
As communication security becomes more and more a problem, when a message is
received, it must authenticated, by using a secret enciphering key. Sometimes, though,
it can fall into wrong hands, thus allowing an unauthentic message to appear to be
authentic. Assume that 80% of all messages received are authentic. Furthermore,
assume that only 5% of all unauthentic messages are sent using the correct key and
that all authentic messages are sent using the correct key. Find the probability that
a message is authentic given that the correct key is used.
How do I compute this event, I've been scratching my head for the last half an hour and tried to look at the problem from different angles, I tried thinking of it as $ P(A/K) $ $A$ - message is authentic and $K$ - key is correct and then tried to compute with Bayes but it makes no sense since $P(K)$ is always going to be 1 since it is specified that the identified key is authentic and $P(K/A)$ would also result in 1, since the message is authentic the key will always be correct. Am I even making sense? I feel like I'm missing something right under my nose because at first glance it doesn't seem to be such a difficult problem, so I would appreciate some help...
probability conditional-probability
asked Feb 2 at 2:49
The VirtuosoThe Virtuoso
448
$endgroup$
add a comment |
$begingroup$
As communication security becomes more and more a problem, when a message is
received, it must authenticated, by using a secret enciphering key. Sometimes, though,
it can fall into wrong hands, thus allowing an unauthentic message to appear to be
authentic. Assume that 80% of all messages received are authentic. Furthermore,
assume that only 5% of all unauthentic messages are sent using the correct key and
that all authentic messages are sent using the correct key. Find the probability that
a message is authentic given that the correct key is used.
How do I compute this event, I've been scratching my head for the last half an hour and tried to look at the problem from different angles, I tried thinking of it as $ P(A/K) $ $A$ - message is authentic and $K$ - key is correct and then tried to compute with Bayes but it makes no sense since $P(K)$ is always going to be 1 since it is specified that the identified key is authentic and $P(K/A)$ would also result in 1, since the message is authentic the key will always be correct. Am I even making sense? I feel like I'm missing something right under my nose because at first glance it doesn't seem to be such a difficult problem, so I would appreciate some help...
probability conditional-probability
asked Feb 2 at 2:49
The VirtuosoThe Virtuoso
448
$endgroup$
As communication security becomes more and more a problem, when a message is
received, it must authenticated, by using a secret enciphering key. Sometimes, though,
it can fall into wrong hands, thus allowing an unauthentic message to appear to be
authentic. Assume that 80% of all messages received are authentic. Furthermore,
assume that only 5% of all unauthentic messages are sent using the correct key and
that all authentic messages are sent using the correct key. Find the probability that
a message is authentic given that the correct key is used.
How do I compute this event, I've been scratching my head for the last half an hour and tried to look at the problem from different angles, I tried thinking of it as $ P(A/K) $ $A$ - message is authentic and $K$ - key is correct and then tried to compute with Bayes but it makes no sense since $P(K)$ is always going to be 1 since it is specified that the identified key is authentic and $P(K/A)$ would also result in 1, since the message is authentic the key will always be correct. Am I even making sense? I feel like I'm missing something right under my nose because at first glance it doesn't seem to be such a difficult problem, so I would appreciate some help...
probability conditional-probability
probability conditional-probability
asked Feb 2 at 2:49
The VirtuosoThe Virtuoso
448
asked Feb 2 at 2:49
The VirtuosoThe Virtuoso
448
asked Feb 2 at 2:49
The VirtuosoThe Virtuoso
448
asked Feb 2 at 2:49
The VirtuosoThe Virtuoso
448
asked Feb 2 at 2:49
The VirtuosoThe Virtuoso
448
448
add a comment |
add a comment |
1 Answer
1
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$begingroup$
Let $A equiv text{{Message is authentic}}, K equiv text{{Message sent with correct key}}$.
You want $$P(A|K) = frac{P(K|A)P(A)}{P(K)}$$
You are correct that $P(K|A) = 1$ because $A subseteq K$. However $P(K) neq 1$. To find $P(K)$ we can use the law of total probability.
Note that $A$ and $overline{A}$ partition the sample space. Then we have that
begin{align}
P(K) &= P(K|A) P(A) + P(K|overline{A}) P(overline{A})\
&= 1cdot 0.80 + 0.5cdot0.20\
&= 0.90
end{align}
Then finally we have
$$P(K) = 8/9$$
answered Feb 2 at 4:06
AlexAlex
52538
$endgroup$
$begingroup$
Thank you, I don't know why but the answer is supposed to be 0.987 but your response really makes sense so...
$endgroup$
– The Virtuoso
Feb 2 at 4:17
add a comment |
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1 Answer
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active
oldest
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Let $A equiv text{{Message is authentic}}, K equiv text{{Message sent with correct key}}$.
You want $$P(A|K) = frac{P(K|A)P(A)}{P(K)}$$
You are correct that $P(K|A) = 1$ because $A subseteq K$. However $P(K) neq 1$. To find $P(K)$ we can use the law of total probability.
Note that $A$ and $overline{A}$ partition the sample space. Then we have that
begin{align}
P(K) &= P(K|A) P(A) + P(K|overline{A}) P(overline{A})\
&= 1cdot 0.80 + 0.5cdot0.20\
&= 0.90
end{align}
Then finally we have
$$P(K) = 8/9$$
answered Feb 2 at 4:06
AlexAlex
52538
$endgroup$
$begingroup$
Thank you, I don't know why but the answer is supposed to be 0.987 but your response really makes sense so...
$endgroup$
– The Virtuoso
Feb 2 at 4:17
add a comment |
$begingroup$
Let $A equiv text{{Message is authentic}}, K equiv text{{Message sent with correct key}}$.
You want $$P(A|K) = frac{P(K|A)P(A)}{P(K)}$$
You are correct that $P(K|A) = 1$ because $A subseteq K$. However $P(K) neq 1$. To find $P(K)$ we can use the law of total probability.
Note that $A$ and $overline{A}$ partition the sample space. Then we have that
begin{align}
P(K) &= P(K|A) P(A) + P(K|overline{A}) P(overline{A})\
&= 1cdot 0.80 + 0.5cdot0.20\
&= 0.90
end{align}
Then finally we have
$$P(K) = 8/9$$
answered Feb 2 at 4:06
AlexAlex
52538
$endgroup$
$begingroup$
Thank you, I don't know why but the answer is supposed to be 0.987 but your response really makes sense so...
$endgroup$
– The Virtuoso
Feb 2 at 4:17
add a comment |
$begingroup$
Let $A equiv text{{Message is authentic}}, K equiv text{{Message sent with correct key}}$.
You want $$P(A|K) = frac{P(K|A)P(A)}{P(K)}$$
You are correct that $P(K|A) = 1$ because $A subseteq K$. However $P(K) neq 1$. To find $P(K)$ we can use the law of total probability.
Note that $A$ and $overline{A}$ partition the sample space. Then we have that
begin{align}
P(K) &= P(K|A) P(A) + P(K|overline{A}) P(overline{A})\
&= 1cdot 0.80 + 0.5cdot0.20\
&= 0.90
end{align}
Then finally we have
$$P(K) = 8/9$$
answered Feb 2 at 4:06
AlexAlex
52538
$endgroup$
Let $A equiv text{{Message is authentic}}, K equiv text{{Message sent with correct key}}$.
You want $$P(A|K) = frac{P(K|A)P(A)}{P(K)}$$
You are correct that $P(K|A) = 1$ because $A subseteq K$. However $P(K) neq 1$. To find $P(K)$ we can use the law of total probability.
Note that $A$ and $overline{A}$ partition the sample space. Then we have that
begin{align}
P(K) &= P(K|A) P(A) + P(K|overline{A}) P(overline{A})\
&= 1cdot 0.80 + 0.5cdot0.20\
&= 0.90
end{align}
Then finally we have
$$P(K) = 8/9$$
answered Feb 2 at 4:06
AlexAlex
52538
answered Feb 2 at 4:06
AlexAlex
52538
answered Feb 2 at 4:06
AlexAlex
52538
answered Feb 2 at 4:06
AlexAlex
52538
52538
$begingroup$
Thank you, I don't know why but the answer is supposed to be 0.987 but your response really makes sense so...
$endgroup$
– The Virtuoso
Feb 2 at 4:17
add a comment |
$begingroup$
Thank you, I don't know why but the answer is supposed to be 0.987 but your response really makes sense so...
$endgroup$
– The Virtuoso
Feb 2 at 4:17
$begingroup$
Thank you, I don't know why but the answer is supposed to be 0.987 but your response really makes sense so...
$endgroup$
– The Virtuoso
Feb 2 at 4:17
$begingroup$
Thank you, I don't know why but the answer is supposed to be 0.987 but your response really makes sense so...
$endgroup$
– The Virtuoso
Feb 2 at 4:17
add a comment |
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