- Step 1: List constraints. A not History. B not Math. C not English. Each chooses a different subject.
- Step 2: Assign subjects. Subjects: History, Math, Science, English. B not Math, so B has History, Science, or English. C not English, so C has History, Math, or Science. A not History, so A has Math, Science, or English.
- Step 3: Test Science. Assume A has Science. Then: B has History or English, C has History or Math, D gets the remaining subject.
- Step 4: Assign remaining. If B has History, C has Math (not English), D has English. Check: A-Science, B-History, C-Math, D-English. Constraints: A not History (satisfied), B not Math (satisfied), C not English (satisfied).
- Step 5: Verify other possibilities. If B has English, C has Math, D has History, A has Science. Valid. Science consistently goes to A.
- Step 6: Final conclusion. Option (1) A is the correct answer.

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: