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1. Let A = { a , b , c ,d , e } and B={ a , b , c , d , e , f , g , h }then A - B is1) A2) B3) BA 4) 2. A Relation R on a Set A is called a partial order, if (A, R) is1) Reflexive relation2) Symmetric relation3) Reflexive, Anti-Symmetric and Transitive relation4) Reflexive, Symmetric and Transitive relation3. If A and B are two independent events such that P(A) =0.5 and   8.0 BAP thenP(B) is1) 0.62) 0.53) 0.84) 0.054. A Context – free grammar G is ambiguous if there is some string w belongs to L(G)that has two distinct1) Graph only2) Parse trees3) Grammars4) Ordered

Question

  1. Let A = { a , b , c ,d , e } and B={ a , b , c , d , e , f , g , h }then A - B is1) A2) B3) BA 4) 2. A Relation R on a Set A is called a partial order, if (A, R) is1) Reflexive relation2) Symmetric relation3) Reflexive, Anti-Symmetric and Transitive relation4) Reflexive, Symmetric and Transitive relation3. If A and B are two independent events such that P(A) =0.5 and   8.0 BAP thenP(B) is1) 0.62) 0.53) 0.84) 0.054. A Context – free grammar G is ambiguous if there is some string w belongs to L(G)that has two distinct1) Graph only2) Parse trees3) Grammars4) Ordered
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Solution

Sure, here are the answers to your questions:

  1. A - B is the set of elements that are in A but not in B. Since all elements of A are also in B, A - B is an empty set. So, the answer is 4) .

  2. A relation R on a set A is called a partial order if it is reflexive, anti-symmetric, and transitive. So, the answer is 3) Reflexive, Anti-Symmetric and Transitive relation.

  3. If A and B are two independent events, the probability of A union B is equal to the sum of the probabilities of A and B. Given that P(A) = 0.5 and P(A U B) = 0.8, we can find P(B) by subtracting P(A) from P(A U B). So, P(B) = P(A U B) - P(A) = 0.8 - 0.5 = 0.3. None of the provided options match this result.

  4. A context-free grammar G is ambiguous if there is some string w that belongs to L(G) that has two distinct parse trees. So, the answer is 2) Parse trees.

This problem has been solved

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