A UMBC student has two different email accounts. Suppose that 70% of her messages come into Account 1. Of the messages that come into Account 1.5% are spam. Meanwhile, of the messages that come into Account 2.8% are spam.

a) If a message is randomly selected, what is the probability that it is NOT a spam?
b) If a randomly selected message is a spam, what is the probability that it came from Account 1.
c) Suppose 5 messages are randomly selected. What is the probability that exactly 2 of them are spam? dy
d) Suppose 20 messages are randomly selected. What is the probability that at least 3 of them are not spam?

a) 0.941

b) 0.593

c) 0.02901

d) 0.9999999999

Step-by-step explanation:

A student has two accounts.

70% of her messages come from account 1.

Probability that a message is on account 1 = P(A) = 0.7

Meaning 30% of her messages come from account 2.

Probability that a message is on account 2 is P(B) = 0.3

Of the messages that come into Account 1, 5% are spam

Calculating in terms of total percentage of messages.

The amount of spam on account 1 is 5% of 70% of the total amount of messages.

P(A n S) = 0.05 × 0.7 = 0.035 = 3.5%

The amount of spam on account 2 is 8% of 70% of the total amount of messages.

P(B n S) = 0.08 × 0.3 = 0.024 = 2.4%

Total amount of spam messages received = P(S) = 2.4% + 3.5% = 5.9% = 0.059

a) If a message is randomly selected, what is the probability that it is NOT a spam?

P(S') = 1 - P(S) = 1 - 0.059 = 0.941

b) If a randomly selected message is a spam, what is the probability that it came from Account 1.

P(A|S) = P(A n S)/P(S)

P(A|S) = (0.035/0.059)

P(A|S) = 0.593

c) Suppose 5 messages are randomly selected. What is the probability that exactly 2 of them are spam

This is a binomial distribution problem

Binomial distribution function is represented by

P(X = x) = ⁿCₓ pˣ qⁿ⁻ˣ

n = total number of sample spaces = 5 messages picked

x = Number of successes required = number of spam messages that should be amongst the 5 messages picked = 2

p = probability of success = probability that a message is a spam message = 0.059

q = probability of failure = probability that a message is NOT a spam message = 0.941

P(X=2) = ⁵C₂ (0.059)² (0.941)⁵⁻²

P(X=2) = 0.02901

d) Suppose 20 messages are randomly selected. What is the probability that at least 3 of them are not spam?

This is also a binomial distribution problem

Binomial distribution function is represented by

P(X = x) = ⁿCₓ pˣ qⁿ⁻ˣ

n = total number of sample spaces = number of messages = 20

x = Number of successes required = at least 3

p = probability of success = probability that a message is not a spam = 0.941

q = probability of failure = probability that a message is a spam message = 0.059

P(X ≥ 3) = 1 - P(X < 3)

P(X < 3) = P(X=0) + P(X=1) + P(X=2) = 0.0000000000001

P(X ≥ 3) = 1 - P(X < 3) = 1 - 0.0000000000001 = 0.9999999999 (Almost 1!)

Hope this Helps!!!

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Step-by-step explanation: