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AURORKA [14]
3 years ago
15

Based on experience, the probability that an incoming email is spam (an undesired message of solicitation) is 38%. If three emai

ls are in my inbox, what is the probability that:A. All three messages are spam.B. At least one of the three messages are spam.C. None of the three messages are spam.
Mathematics
1 answer:
SCORPION-xisa [38]3 years ago
7 0

Answer:

0.0549

0.7617

0.2383

Step-by-step explanation:

Given that:

Probability that incoming email is spam ;

P(spam) = 0.38

If three emails are in my inbox, what is the probability that:

A. All three messages are spam.

P(1st message being spam) * p(second message being spam) * p(3rd message being spam)

= (0.38) * (0.38) * (0.38)

= 0.054872

= 0.0549

B. At least one of the three messages are spam.

1 - p(none of the 3 messages are spam)

P(none of the 3 messages are spam)

P(not spam) = 1 - p(spam) = 1 - 0.38 = 0.62

1 - [P(not spam) * p(not spam) * p(not spam)]

1 - (0.62 * 0.62 * 0.62)

= 1 - 0.238328

= 0.761672

= 0.7617

C. None of the three messages are spam.

P(not spam) = 1 - p(spam) = 1 - 0.38 = 0.62

P(not spam) * p(not spam) * p(not spam)

(0.62 * 0.62 * 0.62)

= 0.238328

= 0.2383

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Answer:

Step-by-step explanation:

Given that Prof  Liu gave the same quiz to the students in her morning class and in her afternoon class. the average score for the two classes combined was 84.

(1) The average score for the students in the morning class was 80

(2) the average score for the students in the afternoon class was 86

Let m be the no of students in I class and n in II class

Total of I class = 80m and

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Total of both classes = 80m +86n

Average =\frac{80m+86n}{m+n} = 84

m cannot be equal to n because in that case average would have been 83

If m >n, then say n =m+a

then average = \frac{80m+86m+86a}{2m+a} =84

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So I class is getting less score of 80 to get a score average of more than 83 (the average) the class II should have more students.

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In a chemical plant, 24 holding tanks are used for final product storage. Four tanks are selected at random and without replacem
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Answer:

a) P(A) = 0,4607     or   P(A) = 46,07 %

b) P(B) = 0,7120   or 71,2 %

c) P(C) = 0,2055  or P(C) = 20,55 %

Step-by-step explanation:

We will use two concepts in solving this problem.

1.- The probability of an event (A) is for definition:

P(A) = Number of favorable events/ Total number of events FE/TE

2.- If A and B are complementary events ( the sum of them is equal to 1) then:

P(A) = 1 - P(B)

a) The total number of events is:

C ( 24,4) = 24! / 4! ( 24 - 4 )!    ⇒  C ( 24,4) = 24! / 4! * 20!

C ( 24,4) = 24*23*22*21*20! / 4! * 20!  

C ( 24,4) = 24*23*22*21/4*3*2

C ( 24,4) = 24*23*22*21/4*3*2    ⇒  C ( 24,4) =  10626

TE = 10626

Splitting the group of tanks in two 6 with h-v  and 24-6 (18) without h-v

we get that total number of favorable events is the product of:

FE = 6* C ( 18, 3)  = 6 * 18! / 3!*15!  =  18*17*16*15!/15!

FE =  4896

Then P(A) ( 1 tank in the sample contains h-v material is:

P(A) = 4896/10626

P(A) = 0,4607     or   P(A) = 46,07 %

b) P(B) will be the probability of at least 1 tank contains h-v

P(B) = 1 - P ( no one tank with h-v)

Again Total number of events is 10626

The total number of favorable events for the ocurrence of P is C (18,4)

FE = C (18,4) = 18! / 14!*4! = 18*17*16*15*14!/14!*4!

FE = 18*17*16*15/4*3*2  = 3060

Then P = 3060/10626

P = 0,2879

And the probability we are looking for is

P(B) = 1 - 0,2879

P(B) = 0,7120   or 71,2 %

c) We call P(C) the probability of finding exactly 1 tank with h-v and t-i

having 4 with t-i tanks is:

reasoning the same way but now having 4 with t-i (impurities) number of favorable events is:

FE = 6*4* C(14,2) = 24 * 14!/12!*2!

FE = 24* 14*13*12! / 12!*2

FE = 24*14*13/2    ⇒  FE = 2184

And again as the TE = 10626

P(C) = 2184/ 10626

P(C) = 0,2055  or P(C) = 20,55 %

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