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It is estimated that 50% of emails are spam emails. Some software has beenapplied to filter these spam emails before they reach your inbox. A certain brandof software claims that it can detect 99% of spam emails, and the probabilityfor a false positive (a non-spam email detected as spam) is 5%.Now if an email is detected as spam, then what is the probability that it isin fact a non-spam email?

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Tundexi

Answer:

P(Non-Spam/Spam Detected) = 5/99

Step-by-step explanation:

To determine the probability that it is in fact a non-spam email, we will employ the use of Bayes Theorem: P(A/B) = (P(B/A) x P(A)) / P(B).

P(Spam) = P(Non-Spam) = 0.5  

P(Spam Detected) = P(Spam) * rate of detection = 0.5 * 0.99

P(Detected/Non-Spam) = P(false positives) = 0.05

Using Bayes Theorem, P(Non-Spam/Spam Detected) = ( P(Spam Detected/Non-Spam) x P(Non-Spam) ) / P(Spam Detected)

= 0.05*0.5 / 0.5*0.99

= 5/99

Using conditional probability, it is found that there is a 0.0481 = 4.81% probability that it is in fact a non-spam email.

Conditional Probability

[tex]P(B|A) = \frac{P(A \cap B)}{P(A)}[/tex]

In which

  • P(B|A) is the probability of event B happening, given that A happened.
  • [tex]P(A \cap B)[/tex] is the probability of both A and B happening.
  • P(A) is the probability of A happening.

In this problem:

  • Event A: Email detected as spam.
  • Event B: It is a non-spam email.

The percentages of emails detected as spam are:

  • 99% of 50%(spam emails).
  • 5% of 50%(non-spam emails).

Hence:

[tex]P(A) = 0.5(0.99) + 0.5(0.05) = 0.52[/tex]

The probability of an email being detected as spam and not being spam is:

[tex]P(A \cap B) = 0.5(0.05)[/tex]

Hence, the conditional probability is:

[tex]P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0.5(0.05)}{0.52} = 0.0481[/tex]

0.0481 = 4.81% probability that it is in fact a non-spam email.

A similar problem is given at https://brainly.com/question/14398287

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