Compute the probability of spam emails given that it does not contain the word "promotion’. (iv) (v) Are both events independent? Explain.

College Algebra
1st Edition
ISBN:9781938168383
Author:Jay Abramson
Publisher:Jay Abramson
Chapter9: Sequences, Probability And Counting Theory
Section9.7: Probability
Problem 5SE: The union of two sets is defined as a set of elements that are present in at least one of the sets....
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Im looking for iv - v. Thanks

Ahmad received 100 emails. 30% of the total emails is spam emails and 70% is
desired emails. The percentage of the word 'promotion' that occurs in the spam
emails is 80%. While 10% in the desired emails contains the word 'promotion'.
(a)
(i)
Draw the tree diagram for the above problem.
(ii)
Compute the probability which contains the word 'promotion'.
(iii)
Compute the probability of desired emails and contains the word
'promotion'.
Compute the probability of spam emails given that it does not contain
the word "promotion'.
(iv)
(v)
Are both events independent? Explain.
Transcribed Image Text:Ahmad received 100 emails. 30% of the total emails is spam emails and 70% is desired emails. The percentage of the word 'promotion' that occurs in the spam emails is 80%. While 10% in the desired emails contains the word 'promotion'. (a) (i) Draw the tree diagram for the above problem. (ii) Compute the probability which contains the word 'promotion'. (iii) Compute the probability of desired emails and contains the word 'promotion'. Compute the probability of spam emails given that it does not contain the word "promotion'. (iv) (v) Are both events independent? Explain.
Expert Solution
Step 1

Given:

P (S) = 0.30

P (D) = 0.70

P (P| S) = 0.80

P (P | D) = 0.10

Step 2

P (P) = PP|S×PS+PP|D×PD            =0.80×0.30+0.10×0.70            =0.31

Tree diagram

Probability homework question answer, step 2, image 1

iv)

Probability of spam email given it does not have word 'promotion'

P (S |NP)

PNP=1-PP            =1-0.31            =0.69PS|NP=PNP|S×PSPNP                =0.20×0.300.69                =0.0870

The required probability is 0.0870

 

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