CS代写 Question 1, Part 1, a

Question 1, Part 1, a
􏰀 P(Rr,Ll,t,Ss,Qq) r,l,q,s
0.9 0.1 +P(¬l|Rr)􏰁P(q|¬l)P(t|q,Rr)+P(¬q|¬l)P(t|¬q,Rr)􏰂􏰄
􏰈 􏰇􏰆 􏰉 􏰈 􏰇􏰆 􏰉

Question 1, Part 1, b Step 1: get equation
P(t) = 􏰀 P(Rr,Ll,t,Ss,Qq) r,l,q,s
= 􏰀 P(Rr)P(Ll|Rr)P(Qq|Ll)P(t|Qq,Rr)P(Ss|Qq,t) r,l,q,s
= 􏰀P(Rr)􏰀P(Ll|Rr)􏰀P(Qq|Ll)P(t|Qq,Rr)􏰀P(Ss|Qq,t) rlqs
(Try to push the summations as far right as you can)
= 􏰀P(Rr)􏰀P(Ll|Rr)􏰀P(Qq|Ll)P(t|Qq,Rr)􏰀P(Ss|Qq,t)
r 􏰈 􏰇􏰆 􏰉 l 􏰈 􏰇􏰆 􏰉 q 􏰈 􏰇􏰆 􏰉􏰈 􏰇􏰆 f1 (R) f2 (L,R) f3 (Q,L) f4 (Q,R)
Step 2: work out factors
(Label the different parts)
S Q f5(S,Q) TT0.9
F T 0.1 FF0.6
(We’ll name the new factor f6)
Notice that we got 1 in both cases. Here, just like in the enu- meration section, we could have just immediately set 􏰅s f5(S, Q) to 1
(We’ll name the new factor f7)
Q R L f7(Q,R,L) TTT 0.63 TTF 0.49 TFT 0.81 TFF 0.63
→ FTT 0.02 FTF 0.06 FFT 0.03 FFF 0.09
􏰉 s 􏰈 􏰇􏰆 􏰉 f5 (S,Q)
L R f2(L,R) Q L f3(Q,L)
TT0.8 TT0.9 TT0.7 T F 0.3 T F 0.7 T F 0.9 F T 0.2 F T 0.1 F T 0.2 FF0.7 FF0.3 FF0.3
Q R f4(Q,R)
R f1(R) T 0.2 F 0.8
Step 3: sum over S
P (t) = 􏰀 f1(R) 􏰀 f2(L, R) 􏰀 f3(Q, L)f4(Q, R) 􏰀 f5(S, Q) rlqs
S Q f5(S,Q) TT 0.9
0.1 → F 0.4+0.6 → F 1 0.6
T 0.9 + 0.1
Step 4: multiply factors
P (t) = 􏰀 f1(R) 􏰀 f2(L, R) 􏰀 f3(Q, L)f4(Q, R)f6(Q) r l q􏰈 􏰇􏰆 􏰉
f3 (Q, L) 0.9 0.7 0.1 0.3 f4(Q,R) 0.7 0.9 0.2 0.3
f7 (Q,R,L)
Q R L f3×f4×f6 T T T 0.9×0.7×1
T T F 0.7×0.7×1 T F T 0.9×0.9×1 T F F 0.7×0.9×1 F T T 0.1×0.2×1 F T F 0.3×0.2×1 F F T 0.1×0.3×1 F F F 0.3×0.3×1
F F QR TT TF FT FF
Q f6(Q) T1 F1

Step 5: sum over Q
Q R L f7(Q,R,L) TTT 0.63 TTF 0.49 TFT 0.81 TFF 0.63 FTT 0.02 FTF 0.06 FFT 0.03 FFF 0.09
Step 6: multiply factors
T T →T F F T F F
f8(R, L) 0.63+0.02
0.49+0.06 0.81+0.03 0.63+0.09
T T 0.65 T F 0.55 F T 0.84 F F 0.72
(We’ll name the new factor f9)
P (t) = 􏰀 f1(R) 􏰀 f2(L, R) 􏰀 f7(Q, R, L) rlq
(We’ll name the new factor f8)
L R f2(L,R) TT 0.8 TF 0.3
R L f8(R,L) → T T 0.65
T T 0.8×0.65
T F 0.2×0.55
F T 0.3×0.84 → F F 0.7×0.72
about the order! In the f2 table, L is on the left. In the other table, L is on the right!
P (t) = 􏰀 f1(R) 􏰀 f2(L, R)f8(R, L) r l􏰈􏰇􏰆􏰉
R L f9 T T 0.52 T F 0.11 F T 0.252 F F 0.504
be careful
Step 7: almost there! Sum over L
R L f9 T T 0.52 T F 0.11 F T 0.252 F F 0.504
T 0.52+0.11 → F 0.252 + 0.504
f10(R) 0.63 0.756
f1 × f10 0.126 0.6048
P (t) = 􏰀 f1(R) 􏰀 f9(R, L) rl
(We’ll name the new factor f10)
Step 8: multiply one more time!
R f1(R) T 0.2
F 0.8 R f10(R) T 0.63 F 0.756
R f1 × f10
→ T 0.2×0.63 → T
P (t) = 􏰀 f1(R)f10(R) r 􏰈 􏰇􏰆 􏰉
(We’ll name the new factor f11)
F 0.8 × 0.756

Step 9: finally, sum over R
Question 1, Part 2 Step 1: Use Bayes rule
Before going on
Step 3: factors
T Q f1(Q,T)
F T 0.1 F T 0.4 F 0.7 F T 0.1
P(t) = 􏰀f11(R) r
= 0.126 + 0.6048 = 0.7308
P(q | s,¬r) = P(q,s,¬r) P(s,¬r)
From here, if we wanted to, we could work out the numerator and denominator individually by following the same process as in the last question. That’s a lot of work, though! It would be better if we did something a bit sneaky instead!
What is that, you ask? Well, let’s think about this for a second. We also know:
P(¬q | s,¬r) = P(¬q,s,¬r) P(s,¬r)
Now, if we knew P (q, s, ¬r) AND P (¬q, s, ¬r), we could work out the denominator by adding them together. Then, we could divide P (q, s, ¬r) by the denominator to get our answer. In other words we could get P(Q, s, ¬r) and normalise this to get our answer. And that’s what we’re going to do!
Step 2: get equation
P(Q,s,¬r) = 􏰀P(Q,Tt,Ll,¬r,s,) t,l
= 􏰀 P (¬r)P (Ll|¬r)P (Q|Ll)P (Tt|Q, ¬r)P (s|Q, Tt) t,l
= P (¬r) 􏰀 P (Tt|Q, ¬r)P (s|Q, Tt) 􏰀 P (Ll|¬r)P (Q|Ll) tl
(Note: there is more than one way to do this! Eg. t and l swapped)
P(Q,s,¬r) = P(¬r)􏰀P(Tt|Q,¬r)P(s|Q,Tt)􏰀P(Ll|¬r)P(Q|Ll)
􏰈􏰇􏰆􏰉 t 􏰈 􏰇􏰆 0.8 f1(T,Q)
􏰉􏰈 􏰇􏰆 􏰉 l 􏰈 􏰇􏰆 􏰉􏰈 􏰇􏰆 􏰉 f2(Q,T ) f3(L) f4(Q,L)
Q T f2(Q,T) T T 0.9
L f3(L) T 0.3
Q L f4(Q,L) T T 0.9
FF0.7 FF0.1

Step 4: multiply
P(Q,s,¬r) = 0.8􏰀f1(T,Q)f2(Q,T)􏰀f3(L)f4(Q,L)
T T T F F T F F
l 􏰈 􏰇􏰆 􏰉 f5 (Q,L)
f5(Q,L) 0.9×0.3 = 0.27
0.7×0.7 = 0.49 0.1×0.3=0.03 0.3×0.7=0.21
Step 5: sum over L
P(Q,s,¬r) = 0.8􏰀f1(T,Q)f2(Q,T)􏰀f5(Q,L) tl
0.27 + 0.49 = 0.76 0.03 + 0.21 = 0.24
Step 6: multiply
Step 7: sum over T
P(Q,s,¬r) = 0.8􏰀f1(T,Q)f2(Q,T)f6(Q) t􏰈 􏰇􏰆 􏰉
Q T f7(Q,T)
T T 0.9×0.9×0.76=0.6156 T F 0.1×0.6×0.76=0.0456 F T 0.3×0.4×0.24=0.0288 F F 0.7×0.1×0.24=0.0168
P(Q,s,¬r) = 0.8􏰀f7(Q,T) t
0.6156 + 0.0456 = 0.6612 0.0288 + 0.0168 = 0.0456

Step 8: normalise
All right, we’ve got:
P(Q, s, ¬r) = 0.8× < 0.6612, 0.0456 >
=< 0.8 × 0.6612, 0.8 × 0.0456 >
=< 0.52896, 0.03648 >
(Note: this 0.8 doesn’t really matter, because we’re going to normalise later anyway)
Now, we’re going to normalise by dividing each term by their sum (ie. 0.52896 + 0.03648), and then we can get our answer!
P(Q | s, ¬r) = < 0.52896, 0.03648 > 0.52896 + 0.03648
= < 0.52896, 0.03648 > 0.56544
= < 0.52896, 0.03648 > 0.56544
raining is about 0.935!
Question 1, Part 3
≈< 0.935, 0.065 >
∴ P (q | s, ¬r) ≈ 0.935 (∴ means “therefore”)
(≈ means “approximately”) So, the probability of quokkas being happy given that people are taking lots of quokka selfies and it is not
Well, in the previous question we got a probability of 0.935 that the quokkas are happy given the evidence. So, since this is more than 0.5, we will predict that they are happy. Yay! 😀 (I guess you could say we BAYESically had the answer already… :D)
Question 1, Parts 4 and 5
P(r | ¬l,s) = 0.0619 P(l | q,t,s) = 0.44218

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