Roger Federer gets a chance of 60% to win 1st set of tennis and Andy Murray gets a chance of 55% to win 2nd set of tennis. In what Percent of cases are they likely to contradict each other, narrating the same incident?
Let ‘A’ be the event that Roger Federer wins 1st set. Let ‘B’ be the event that Andy Murray wins 2nd set. Then, A’ = Event that the Roger Federer losses 1st set And B’ = Event that the Andy Murray losses 2nd set. Therefore, P(A) =60/100 = 12/20 P(B) = 55/100 = 11/20 P(A’) = 1 – (12/20) = 8/20 P(B’) = 1 – (11/20) = 9/20 First, we have to find the probability that they contradict each other. That is, P(A and B contradicts each other) = P[(Roger Federer win in 1st set and Andy Murray losses in 2nd set) (or) (Roger Federer losses in 1st set and Andy Murray wins in 2nd set)] =P(A) × P(B’) + P(A’) × P(B) = 12/20  ×  9/20+  11/20  ×  8/20 = 196/400 Required Percentage = 196/400  ×100 = 49%
Statements: F @ R, R $ J, V % J, V # Z
Conclusions: I. F * VÂ Â Â Â Â Â Â II. R * VÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â...
Statement: L = C; E ≥ M ≥ U ≥ C
Conclusion:
I. E > L
II. E = L
Statement: W>Y<X<Z=U>S; W<T ≥V
I. Y<T
II. X > V
Statements : P > Q < R = U ≤ V = S ≤ W ≥ X > I
Conclusions :
I. Q ≥ V
II. R ≤ W
Statements: Â A % B & G % B; B # L & J; J @ K # S
Conclusions:
I. L @ K
II. A % K
III. S @ B
...Statements: A > C ≥ B = D; E < F = G < H = I ≤ D
Conclusions:
I. B > E
II. D < E
III. E ≥ B
Statements: J $ K, K * T, T @ N, N © R
Conclusions:
 I. J $ T                  II.R * T               Â...
Statements: O< V ≤ N = P < S, R = Z ≥ Y = X > O
Conclusions:
I. R > N
II. X > P
III. R > O
Statement:X=Y ≥ Z > Q; Y < V ; W < Q
Conclusions:
I. V > W
II. Q > V
Statements: A & D, D # P, P @ Q, Q % R
Conclusions: I. D & R                    II. Q # A
...