Shortcut Trick
Total cases for Office P transition: (M→M, M→F, F→M, F→F).
P(M) = [P(Mout)×P(Min)×P(Mfinal)] + [P(Mout)×P(Fin)×P(Mfinal)] + [P(Fout)×P(Min)×P(Mfinal)] + [P(Fout)×P(Fin)×P(Mfinal)]
P(M) = [(6/10×5/11×6/10) + (6/10×6/11×5/10) + (4/10×4/11×7/10) + (4/10×7/11×6/10)]
P(M) = (180 + 180 + 112 + 168) / 1100 = 640 / 1100 = 32/55.
∴ The correct answer is 32/55.
Alternate Method
Given:
Office P: 6 Males, 4 Females (Total = 10)
Office Q: 4 Males, 6 Females (Total = 10)
Let Emm be Male from P to Q, then Male from Q to P.
Let Emf be Male from P to Q, then Female from Q to P.
Let Efm be Female from P to Q, then Male from Q to P.
Let Eff be Female from P to Q, then Female from Q to P.
Calculations:
⇒ P(Emm) = (6/10) × (5/11) = 30/110
⇒ P(Emf) = (6/10) × (6/11) = 36/110
⇒ P(Efm) = (4/10) × (4/11) = 16/110
⇒ P(Eff) = (4/10) × (7/11) = 28/110
Probability of selecting Male from P after these cases:
⇒ P(X/Emm) = 6/10 (Composition stays 6M, 4F)
⇒ P(X/Emf) = 5/10 (Composition becomes 5M, 5F)
⇒ P(X/Efm) = 7/10 (Composition becomes 7M, 3F)
⇒ P(X/Eff) = 6/10 (Composition stays 6M, 4F)
⇒ Total P(X) = (30/110 × 6/10) + (36/110 × 5/10) + (16/110 × 7/10) + (28/110 × 6/10)
⇒ P(X) = (180 + 180 + 112 + 168) / 1100 = 640 / 1100 = 32/55
∴ The correct answer is 32/55.
Additional Information
Law of Total Probability
It relates marginal probabilities to conditional probabilities. P(A) = Σ P(A|Bi)P(Bi) for a partition Bi.
Independent Events
Two events are independent if the occurrence of one does not affect the probability of the other: P(A ∩ B) = P(A) × P(B).
Baye's Theorem
Determines the probability of an event based on prior knowledge: P(A|B) = [P(B|A)P(A)] / P(B).