(a) $P(\text{solved})=1-\prod(1-p_i)=1-\tfrac12\cdot\tfrac23\cdot\tfrac35=\tfrac45$. (b) Network passes give critical path $1\to2\to3\to5\to8$, duration $36$ days.

A 5-mark either/or from the Probability / Operations Research units of the Tamil Nadu HSC Class-11 Business Mathematics & Statistics syllabus. Both alternatives are solved.

(a) Probability the problem is solved.

Step 1 — Probabilities of failure. $X,Y,Z$ solve with $\tfrac12,\tfrac13,\tfrac25$, so they fail with

$1-\tfrac12=\tfrac12,\qquad 1-\tfrac13=\tfrac23,\qquad 1-\tfrac25=\tfrac35.$

Step 2 — Probability none solves (independent events):

$P(\text{none})=\frac12\cdot\frac23\cdot\frac35=\frac{6}{30}=\frac15.$

Step 3 — At least one solves.

$P(\text{solved})=1-P(\text{none})=1-\frac15=\frac45=0.8.$

(b) Network analysis (CPM). Activities and durations: $1\text{-}2(7),\ 1\text{-}6(6),\ 2\text{-}3(14),\ 2\text{-}4(5),\ 3\text{-}5(11),\ 4\text{-}5(7),\ 6\text{-}7(11),\ 5\text{-}8(4),\ 7\text{-}8(18).$

Step 1 — Forward pass (earliest event times $E$).

$E_1=0,\ E_2=7,\ E_3=21,\ E_4=12,\ E_6=6,$

$E_5=\max(E_3+11,\ E_4+7)=\max(32,19)=32,\quad E_7=E_6+11=17,$

$E_8=\max(E_5+4,\ E_7+18)=\max(36,35)=36.$

Step 2 — Backward pass (latest event times $L$).

$L_8=36,\ L_5=32,\ L_7=18,\ L_3=21,\ L_4=25,\ L_6=7,$

$L_2=\min(L_3-14,\ L_4-5)=\min(7,20)=7,\quad L_1=\min(L_2-7,\ L_6-6)=\min(0,1)=0.$

Step 3 — Activity times ($EST=E_i,\ EFT=E_i+d,\ LFT=L_j,\ LST=L_j-d$, Float $=LST-EST$).

| Activity | $d$ | EST | EFT | LST | LFT | Float | …