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
If P(A) = 0.7, P(B) = 0.5 and P(B/A) = 0.3, find (i) P(A/B) (ii) P(A ∪ B)?
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?
20 percent of the pens produced in a factory are of red colour and 4 percent are red and defective. If one pen is picked up at random, then what is the probability of its being defective if it is red?
In a game, there are three rooms- I, Il and IIl. Room I contain 2 boxes having gift items and 3 empty boxes, room II contains 3 boxes having gift items and 2 empty boxes, and room III contains 4 boxes having gift items and one empty box respectively. There is an equal probability of each room being chosen by a player. Mr John selects one box from a room chosen at random. The probability that Mr John wins a box having gift items is:
