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
Q2. What is P(G | F)?
Q3. What is P(H | F)? Compare P(H | F) vs. P(E | F) and P(H | F) vs. P(H) and explain which is larger for each case.
Q4. What is P(F | H)? Compare P(H | F) vs. P(F | H) and explain which is larger.
Q5. What is P(I | E)?
Q6. What is P(I | F)?

Answer 1

Solution-

We have given the following information-

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.

We know the Conditional Probability of happening of event A if it is given that B has happen is

P(A/B)= P(Event A and event B both occurs)÷P(B)

We also know the number of possibilities of getting n mails out of m is n is the combination of n out of m

=C(m,n).

Now,

(2)

P(G | F)

=P(G and F happen)÷P(F)

=[P(user#2, #3 receives 6,5 spam mails respectively, user#3 receives 1 mail out of remaining mails 19 and other receives 6 mails out of remaining mails respectively)] ÷ [P(user #2 received 6 spam mails and other receives 6 mails out if remaining mails respectively)]

=[C(10,6)×C(4,5)×C(19,1)×C(18,6)×C(12,6)×(6,6)]÷

[(10,6)×C(24,6) ×C(18,6)×C(12,6)×P(6,6)]

=[C(4,5)×C(19,1)] ÷ [C(24,6)]

=[0×C(19,1)] ÷ [C(24,6)]

=0

[Since user #2 receives 6 spam mails then then 10-6 =4 spam mails are left.So, to receive 5 mails by user #3 is not possible.So here C(4,5)=0]

Hence, P(G/F)=0

(3)

P(H | F)

=[P( user#2,#4 receives 6, 1 spam mails out of 10 respectively , user #4 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)]

=[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)]

=[C(4,1)×C(20,5)]÷ [C(24,6)]

=(4×1550) ÷ (134596)

=0.460757

Hence, P(H/F)=0.460757

Now,

P(E | F)

=[P( user#2,#1 receives 6, 3 spam mails out of 10 respectively , user #1 receives 3 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)]

=[C(10,6)×C(4,3)×C(20,3)×C(18,6)×C(12,6)×C(6,6)]÷

[C(10,6)×C(24,6)×C(18,6)×C(12,6)×C(6,6)]

=[C(4,3)×C(20,3)] ÷ [C(24,6)]

=(4×1140)÷(134596)

=0.033879

Hence, P(E/F)=0.033879

Hence, P(H/F) is larger than P(E/F).

Now,

P(H)

=P(User #4 receives 1 spam mail out of 10 and 5 non spam mails out of 20)

=[C(10,1)×C(20,5) ] ÷ [C(30,6)]

=(10×15504)÷(593775)

=0.26119

Hence, P(H)=0.26119

Here, also P(H/F)is larger than P(H).

Similarly

(4)

P(F | H)

=[P( user#4,#2 receives 1, 6 spam mails out of 10 respectively , user #4 receives 5 non spam mails out of remaining 20 non spam mails and other users receives 6 mails out remaining respectively) ] ÷ [P(user #4 receives 1 spam mails out of 10 , user #4 receives 5 non spam mails out of 20 and other users receives 6 mails out of remaining respectively)]

=[C(10,1)×C(9,6)×C(20,5)×C(18,6)×C(12,6)×C(6,6)]÷

[C(10,1)×C(20,5)×C(24,6)×C(18,6)×C(12,6)×C(6,6)]

=C(9,6)÷C(24,6)

=84÷134596

=0.000624

Hence, P(F/H)=0.000624

Hence, P(H | F) is larger than P(F | H).

(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

Hence, P(I/E)=0.940711

(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)]

=[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 - 0.460757

=0.539243

Hence, P(I/F)=0.539243