Answer:
a) 0.941
b) 0.593
c) 0.02901
d) 0.9999999999
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!!!