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Let events C1, C2, C3 be the following : C1 : the calculator is manufactured by machine A C2 : the calculator is manufactured by machine B C3 : the calculator is manufactured by machine C Clearly, C1, C2, C3 are mutually exclusive and exhaustive events and hence, they represent a partition of the sample space. Let the event E be ‘the calculator is defective’. The event E occurs with C1 or with C2 or with C3. Given that, P(C1) = 25% = 0.25, P (C2) = 0.35 and P(C3) = 0.40 Again P(E|C1) = Probability that the calculator is defective given that it is manufactured by machine A = 5% = 0.05 Similarly, P(E|C2) = 0.04, P(E|C3) = 0.02. Hence, by Bayes' Theorem, we have P(C2|E) = [P(C2)P(E|C2)]/[P(C1)P(E|C1)+P(C2)P(E|C2)+P(C3)P(E|C3)] => [0.35 × 0.04]/[0.25 × 0.05 + 0.35 × 0.04 + 0.40 × 0.02] => 0.0140/0.0345 = 28/69
Statements: V ≤R = W ≥ Q, U = T ≥ S < X, U < Q
Conclusions: I. V < Q II. Q > X
Statements: Q © E, S % C, E $ S, C @ A
Conclusions:
I. A © C
II. S % A
III. C © Q
Statements: L # W, W % V, V $ H, H # T
Conclusions : I. V @ T II. H & W III .V # T
...Which of the following symbols should be placed in the blank spaces respectively (in the same order from left to right) in order to complete the given e...
Which of the following is true in the given expression?
G < H ≤ I, V ≥ W = G, R ≤ I = A
In which of these expression ‘X > T’ is definitely True?
Statements: Q ≥ R > U; R ≤ S; U ≥ B
Conclusions: I. B < R II. B ≤ Q
Statement: Z > F ≥ O; Z ≤ G = P; Q > F
Conclusion: I. P > O II. Q > G
Statements: Q > S ≥ R = T; U < V = W < X = Y ≤ T
Conclusions:
I. R > U
II. T < U
III. U ≥ R
Statement: T > U ≥ V; T ≤ W = X; I > U
Conclusion: I. U < X II. I > T