The probability that A speaks truth is 4 / 5 while this probability for B is 3 / 4. The probability that they contradict each other when asked to speak on a fact is
7 / 20
Understanding Probability of Contradiction
This problem involves calculating the probability of two independent events occurring in a specific way. We are given the probabilities of two individuals, A and B, speaking the truth, and we need to find the probability that they contradict each other when asked about a fact.
Given Probabilities:
- The probability that A speaks truth is given as $ P(A_{truth}) = \frac{4}{5} $.
- The probability that B speaks truth is given as $ P(B_{truth}) = \frac{3}{4} $.
Calculating Probabilities of Lying:
For A and B to contradict each other, one must speak the truth while the other lies. First, let's find the probabilities of them lying:
- The probability that A lies is $ P(A_{lie}) = 1 - P(A_{truth}) $. $ P(A_{lie}) = 1 - \frac{4}{5} = \frac{5}{5} - \frac{4}{5} = \frac{1}{5} $.
- The probability that B lies is $ P(B_{lie}) = 1 - P(B_{truth}) $. $ P(B_{lie}) = 1 - \frac{3}{4} = \frac{4}{4} - \frac{3}{4} = \frac{1}{4} $.
Identifying Contradiction Scenarios:
Two possible scenarios lead to A and B contradicting each other:
- Scenario 1: A speaks the truth, and B lies.
- Scenario 2: A lies, and B speaks the truth.
Calculating Probability for Each Scenario:
Since the events of A speaking or lying are independent of B speaking or lying, we can multiply their probabilities:
- Probability of Scenario 1 (A speaks truth AND B lies): $ P(A_{truth} \text{ and } B_{lie}) = P(A_{truth}) \times P(B_{lie}) $ $ P(A_{truth} \text{ and } B_{lie}) = \frac{4}{5} \times \frac{1}{4} = \frac{4}{20} $
- Probability of Scenario 2 (A lies AND B speaks truth): $ P(A_{lie} \text{ and } B_{truth}) = P(A_{lie}) \times P(B_{truth}) $ $ P(A_{lie} \text{ and } B_{truth}) = \frac{1}{5} \times \frac{3}{4} = \frac{3}{20} $
Total Probability of Contradiction:
The total probability that they contradict each other is the sum of the probabilities of these two mutually exclusive scenarios:
$ P(\text{contradiction}) = P(A_{truth} \text{ and } B_{lie}) + P(A_{lie} \text{ and } B_{truth}) $
$ P(\text{contradiction}) = \frac{4}{20} + \frac{3}{20} = \frac{4+3}{20} = \frac{7}{20} $
Conclusion:
The probability that A and B contradict each other is $ \frac{7}{20} $. This corresponds to the third option provided.
Important Questions from Multiplication Theorem of Events
For any two events A and B, the probability that at least one of them occur is 0.6. If A and B occur simultaneously with a probability 0.3, then P(A') + P(B') is
A student appears for tests I, II and III. The student is considered successful if the passes in tests I, II or I, III or all the three. The probabilities of the student passing in test I, II and III are m, n and 1/2 respectively. If the probability of the student to be successful is 1/2, then which one of the following is correct?
If \(\rm P(A\cup B)=\dfrac{5}{6}, P(A\cap B)=\dfrac{1}{3}\:and\:P(\bar A)=\dfrac{1}{2}\) , then which of the following is/are correct?
1. A and B are independent events.
2. A and B are mutually exclusive events.
Select the correct answer using the code given below.
In a lottery of 10 tickets numbered 1 to 10, two tickets are drawn simultaneously. What is the probability that both the tickets drawn have prime numbers?
In a series of 3 one-day cricket matches between teams A and B of a college, the probability of team A winning or drawing are 1/3 and 1/6 respectively. If a win, loss or draw gives 2, 0 and 1 point respectively, then what is the probability that team A will score 5 points in the series?
