See image — AITS & Test Series Chemistry Question

💡 Solution & Explanation

P  3, Q  2, R  1, S  4 (P) Required probability = 1 2 . (Q) Required probability 3 4 4 4 4 7 7 1 1 1 1 1 1 47 . 2 2 2 2 2 2 2                                            (R) Required probability 7 4 . 2  (S) There are five cases 0, 1, 3 or 4 wins Required probability 7 7 1 7 15 10 1 34 2 2       .