Two students Anil and Ashima appeared in an examination. The probability that Anil will qualify the examination is 0.05 and that Ashima will qualify the examination is 0.10 . The probability that both will qualify the examination is 0.02 . Find the probability that (a) Both Anil and Ashima will not qualify the examination. (b) Atleast one of them will not qualify the examination and (c) Only one of them will qualify the examination.

Answer: (a) The probability that both Anil and Ashima will not qualify the examination is 0.87. (b) The probability that at least one of them will not qualify the examination is 0.98. (c) The probability that only one of them will qualify the examination is 0.11.

Step-by-step solution

Step 1: Define Events and Given Probabilities

Let AA be the event that Anil qualifies the examination, and BB be the event that Ashima qualifies the examination. We are given the probabilities for these events and the probability that both qualify.

Step 2: Calculate Probability of Union

The probability that at least one of them qualifies is given by the formula for the union of two events. We substitute the given probabilities into this formula.

Step 3: Part (a): Both will not qualify

The event that both Anil and Ashima will not qualify is the complement of the event that at least one of them qualifies. We use De Morgan's law, P(A′∩B′)=P((A∪B)′)P(A' \cap B') = P((A \cup B)'), and the complement rule P(E′)=1−P(E)P(E') = 1 - P(E).

Step 4: Part (b): At least one will not qualify

The event that at least one of them will not qualify is the complement of the event that both qualify. We use De Morgan's law, P(A′∪B′)=P((A∩B)′)P(A' \cup B') = P((A \cap B)'), and the complement rule.

Step 5: Part (c): Only one will qualify

The probability that only one of them qualifies can be found by subtracting the probability that both qualify from the probability that at least one qualifies. Alternatively, it can be calculated as P(A∩B′)+P(A′∩B)P(A \cap B') + P(A' \cap B).

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