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NCERT Exemplar · Q38

Q.If A and B are two candidates seeking admission in an engineering College. The probability that A is selected is .5.5 and the probability that both A and B are selected is at most .3.3. Is it possible that the probability of B getting selected is 0.70.7?

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Yes, it is possible for the probability of B getting selected to be 0.70.7. This is because the given conditions allow for a valid probability of both A and B being selected, specifically in the range 0.2≤P(A∩B)≤0.30.2 \le P(A \cap B) \le 0.3.

In probability, we operate under a set of fundamental rules, known as axioms, that ensure consistency and logical coherence. One crucial axiom states that the probability of any event must be between 00 and 11 (inclusive). Another important derived rule is the Addition Rule for two events, which relates the probabilities of individual events to the probability of their union and intersection.

The question asks if a specific probability for event B is possible given other constraints. To determine possibility, we need to check if assuming P(B)=0.7P(B) = 0.7 leads to any contradiction with the fundamental axioms or the given information. If we can find a scenario where all conditions are met, then it is possible.

Let's define the events:

  • AA: Candidate A is selected.
  • BB: Candidate B is selected.

We are given the following probabilities:

  • P(A)=0.5P(A) = 0.5
  • P(A∩B)≤0.3P(A \cap B) \le 0.3 (The probability that both A and B are selected is at most 0.30.3)

We need to determine if it is possible that P(B)=0.7P(B) = 0.7.

  1. Recall the Addition Rule for Probabilities: For any two events AA and BB, the probability of their union (either A or B or both are selected) is given by:

P(A∪B)=P(A)+P(B)−P(A∩B)P(A \cup B) = P(A) + P(B) - P(A \cap B)

  1. Apply the Axiom of Total Probability: A fundamental axiom of probability states that the probability of any event cannot exceed 11. Therefore, the probability of the union of A and B must also be less than or equal to 11:

P(A∪B)≤1P(A \cup B) \le 1

Combining this with the Addition Rule, we get:

P(A)+P(B)−P(A∩B)≤1P(A) + P(B) - P(A \cap B) \le 1

  1. Substitute the given and hypothetical values: Let's assume P(B)=0.7P(B) = 0.7 and substitute the given P(A)=0.5P(A) = 0.5 into the inequality:

0.5+0.7−P(A∩B)≤10.5 + 0.7 - P(A \cap B) \le 1

1.2−P(A∩B)≤11.2 - P(A \cap B) \le 1

  1. Determine the implied lower bound for P(A∩B)P(A \cap B): Rearranging the inequality to isolate P(A∩B)P(A \cap B):

1.2−1≤P(A∩B)1.2 - 1 \le P(A \cap B)

0.2≤P(A∩B)0.2 \le P(A \cap B)

This means that if $P(B) = 0.7$, then the probability of both A and B being selected must be at least $0.2$. …

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