Of the emails received by an office, 20% are spam. The word “offer” appears in 50% of the spam emails and in 10% of the other emails. An email is chosen at random. Match List-I with List-II
| List-I | List-II | ||
|---|---|---|---|
| (A) | Probability that the email contains the word “offer” | (I) | \(\displaystyle\frac{5}{9}\) |
| (B) | Probability that the email is spam, given that it contains the word | (II) | \(\displaystyle\frac{9}{50}\) |
| (C) | Probability that the email is not spam, given that it contains the word | (III) | \(\displaystyle\frac{5}{41}\) |
| (D) | Probability that the email is spam, given that it does not contain the word | (IV) | \(\displaystyle\frac{4}{9}\) |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → (A)-(II), (B)-(I), (C)-(IV), (D)-(III)
Let \(E_1\) and \(E_2\) be the events that the email is spam and not spam, and W the event that it contains the word.
\(P(E_1)P(W|E_1) = 0.2\times 0.5 = 0.10\)
\(P(E_2)P(W|E_2) = 0.8\times 0.1 = 0.08\)
(A) Probability that the email contains the word “offer”
This matches (II) \(\frac{9}{50}\).
By the theorem of total probability:
\( \displaystyle P(W) = 0.10+0.08 = 0.18 = \frac{9}{50} \)
(B) Probability that the email is spam, given that it contains the word
This matches (I) \(\frac{5}{9}\).
By Bayes’ theorem:
\( \displaystyle P(E_1|W) = \frac{0.10}{0.18} = \frac{5}{9} \)
(C) Probability that the email is not spam, given that it contains the word
This matches (IV) \(\frac{4}{9}\).
\( \displaystyle P(E_2|W) = \frac{0.08}{0.18} = \frac{4}{9} \)
(D) Probability that the email is spam, given that it does not contain the word
This matches (III) \(\frac{5}{41}\).
A spam email lacks the word with probability 0.5, and another email with probability 0.9:
\(P(E_1)P(W'|E_1) = 0.2\times 0.5 = 0.10\)
\(P(E_2)P(W'|E_2) = 0.8\times 0.9 = 0.72\)
\( \displaystyle P(E_1|W') = \frac{0.10}{0.10+0.72} = \frac{0.10}{0.82} = \frac{5}{41} \)