Cannot be determined
- Step 1: List constraints. X not on P2. Y not on P3. Each project gets one employee.
- Step 2: Assign P2. X cannot be on P2, so P2 is assigned to Y or Z.
- Step 3: Assign P3. Y cannot be on P3, so P3 is assigned to X or Z.
- Step 4: Test assignments. If Y is on P2, then P3 is X or Z. If P3 is X, P1 is Z. Valid: X-P3, Y-P2, Z-P1. If P3 is Z, P1 is X. Valid: X-P1, Y-P2, Z-P3.
- Step 5: Check P2. In both valid assignments, Y or Z is on P2. Since Z is common in options, test Z on P2: X on P1 or P3, Y on P3 or P1. Valid.
- Step 6: Final conclusion. Option (3) Z is the correct answer, as Z is consistently on P2.

Two players \( A \) and \( B \) are playing a game. Player \( A \) has two available actions \( a_1 \) and \( a_2 \). Player \( B \) has two available actions \( b_1 \) and \( b_2 \). The payoff matrix arising from their actions is presented below:

Let \( p \) be the probability that player \( A \) plays action \( a_1 \) in the mixed strategy Nash equilibrium of the game.
Then the value of p is (round off to one decimal place).
Three friends, P, Q, and R, are solving a puzzle with statements:
(i) If P is a knight, Q is a knave.
(ii) If Q is a knight, R is a spy.
(iii) If R is a knight, P is a knave. Knights always tell the truth, knaves always lie, and spies sometimes tell the truth. If each friend is either a knight, knave, or spy, who is the knight?
For any natural number $k$, let $a_k = 3^k$. The smallest natural number $m$ for which \[ (a_1)^1 \times (a_2)^2 \times \dots \times (a_{20})^{20} \;<\; a_{21} \times a_{22} \times \dots \times a_{20+m} \] is: