Question

Consider: 30 emails are sent, 6 each to 5 users. 10 of the 30 emails are spam. All possible outcomes are equally likely.
E = user #1 receives 3 spam emails
F = user #2 receives 6 spam emails
G = user #3 receives 5 spam emails
H = user #4 receives 1 spam emails
I = user #5 receives at least 2 spam emails
Q5. What is P(I | E)?
Q6. What is P(I | F)?

Answer 1

Answer:

Given That:

30 emails are sent, 6 each to 5 users. 10 of the 30 emails are spam with all possible outcomes are equally likely and

E = user #1 receives 3 spam emails
F = user #2 receives 6 spam emails
G = user #3 receives 5 spam emails
H = user #4 receives 1 spam emails
I = user #5 receives at least 2 spam emails.

5)

P(I | E)

= [P( user#1,#5 receives 3, 1 spam mails out of 10 respectively , user #1 receives 3 non spam mails out of remaining 20 non spam mails thtn user #5 receive 5 non spam mails out of remaining 17 and other users receives 6 mails out remaining respectively) ] ÷ [P(user #1 receives 3 spam mails out of 10 and 3 no n spam mails out of 20 other users receives 6 mails out of remaining respectively)]

=1-[C(10,3)×C(7,1)×C(20,3)×C(17,1)×C(18,6)×C(12,6)×C(6,6)]÷ [C(10,3)×C(17,1)×C(24,6)×C(18,6)×C(12,6)×C(6,6)]

= 1 -[C(7,1)×C(20,3)]÷[C(24,6)]

= 1 - (7×1140÷134596)

=1- 0.059289

=0.940711

Note:

C(n,r)=?C(n,r)=?

C(n,r)=C(20,3)C(n,r)=C(20,3)

=20!(3!(20−3)!)=20!(3!(20−3)!)

= 1140

and

24!(6!(24−6)!)=24!(6!(24−6)!)

= 134596

6.

P(I | F)

=1- [P( user#2,#5 receives 6, 1 spam mails out of 10 respectively , user #5 receives 5 non spam mails out of remaining 20 non spam mails and other users receives 6 mails out remaining respectively) ] ÷ [P(user #2 receives 6 spam mails out of 10 and other users receives 6 mails out of remaining respectively)]

=1-[C(10,6)×C(4,1)×C(20,5)×C(18,6)×C(12,6)×C(6,6)]÷ [C(10,6)×C(24,6)×C(18,6)×C(12,6)×C(6,6)]

= 1- P(H/F)

= 1-[C(4,1)×C(20,5)÷C(24,6)]

=1 - 0.460757

= 0.539243

Hence, P(I/F)=0.539243