It is observed that 50% of mails are spam. There is a software that filters spam mail before reaching the inbox. Its accuracy for detecting a spam mail is 99% and chances of tagging a non-spam mail as spam mail is 5%. If a certain mail is tagged as spam find the probability that it is not a spam mail.
Spam Mail Probability Calculation Using Bayes' Theorem
This solution explains how to find the probability that a mail tagged as spam is actually not a spam mail, given specific accuracy rates for a spam filtering software.
Defining the Events
Let's define the events involved in this probability problem:
- S: The event that a mail is actually Spam.
- NS: The event that a mail is Not Spam.
- T: The event that a mail is Tagged as Spam by the filter.
Given Probabilities
From the question, we are given the following probabilities:
- The probability that a mail is spam: \( P(S) = 50\% = 0.50 \)
- The probability that a mail is not spam: \( P(NS) = 1 - P(S) = 1 - 0.50 = 0.50 \)
- The accuracy of the filter for detecting a spam mail (True Positive Rate): \( P(T|S) = 99\% = 0.99 \)
- The probability that the filter tags a non-spam mail as spam (False Positive Rate): \( P(T|NS) = 5\% = 0.05 \)
Objective of the Question
We need to find the probability that a mail is not spam given that it has been tagged as spam. In notation, we need to calculate \( P(NS|T) \).
Applying Bayes' Theorem
Bayes' Theorem is the perfect tool for this problem. It helps us update our belief (probability) about an event based on new evidence. The formula is:
$ P(NS|T) = \frac{P(T|NS) \times P(NS)}{P(T)} $To use this formula, we first need to calculate the total probability of a mail being tagged as spam, \( P(T) \).
Calculating the Total Probability of Tagging as Spam (P(T))
A mail can be tagged as spam in two ways:
- It is actually spam and the filter correctly tags it as spam.
- It is not spam, but the filter incorrectly tags it as spam.
The total probability \( P(T) \) is calculated as:
$ P(T) = P(T \cap S) + P(T \cap NS) $Using the definition of conditional probability, \( P(A \cap B) = P(A|B) \times P(B) \), we can rewrite this as:
$ P(T) = P(T|S) \times P(S) + P(T|NS) \times P(NS) $Now, let's plug in the values:
$ P(T) = (0.99 \times 0.50) + (0.05 \times 0.50) $ $ P(T) = 0.495 + 0.025 $ $ P(T) = 0.520 $So, the overall probability that any given mail gets tagged as spam is 0.520.
Calculating the Final Probability P(NS|T)
Now we can substitute the values into Bayes' Theorem:
$ P(NS|T) = \frac{P(T|NS) \times P(NS)}{P(T)} $ $ P(NS|T) = \frac{0.05 \times 0.50}{0.520} $ $ P(NS|T) = \frac{0.025}{0.520} $Simplifying the Result
To express the answer as a simple fraction, we can convert the decimals to fractions or multiply the numerator and denominator by 1000 to remove decimals:
$ P(NS|T) = \frac{0.025 \times 1000}{0.520 \times 1000} = \frac{25}{520} $Now, we simplify this fraction by dividing both the numerator and the denominator by their greatest common divisor, which is 5:
$ P(NS|T) = \frac{25 \div 5}{520 \div 5} = \frac{5}{104} $Conclusion
Therefore, if a certain mail is tagged as spam, the probability that it is actually not a spam mail is \( \frac{5}{104} \).
Important Questions from Conditional Probability
Let A and B be two events such that \(P(\overline{A \cup B}) = \dfrac{1}{6}\) , P(A ∩ B) = \(\dfrac{1}{4}\) and P(A̅) = \(\dfrac{1}{4}\) , where A̅ stands for complement of event A. Then, events A and B are:
A bike manufacturing factory has two plants P and Q. Plant P manufactures 60 percent of bikes and plant Q manufacture 40 percent. 80 percent of the bikes at plant P and 90 percent of the bikes at plant Q are rated of standard quality. A bike is chosen at random and is found to be of standard quality. What is the probability that it has come from plant P?
A and B are two events such that A̅ and B̅ are mutually exclusive. If P(A) = 0.5 and P(B) = 0.6, then what is the value of P(A|B)?
For two dependent events A and B, it is given that P(A) = 0.2 and P(B) = 0.5. If A ⊆ B, then the values of conditional probabilities P(A|B) and P(B|A) are respectively
In a bulb factory, machines P, Q and R manufacture respectively 25%, 35% and 40% of the total. Of their output 5, 4 and 2 percent respectively are defective bulbs. A bulb is drawn at random and it is found to be defective. What is the probability that it was manufactured by machine Q?
