A speaks the truth 3 out o f 4 times, and B 5 out o f 6 times. What is the probability that they will contradict each other in stating the same fact?a)2/3b)1/3c)5/6d)None of theseCorrect answer is option 'B'. Can you explain this answer?

Question

Answer

To solve this problem, we need to find the probability that A and B will contradict each other in stating the same fact.

Given information:
- A speaks the truth 3 out of 4 times, which means A has a probability of 3/4 = 0.75 of speaking the truth.
- B speaks the truth 5 out of 6 times, which means B has a probability of 5/6 ≈ 0.83 of speaking the truth.

We can approach this problem by considering two scenarios:
1. A tells the truth and B lies.
2. A lies and B tells the truth.

Probability of A telling the truth and B lying:
- Probability of A telling the truth = 0.75
- Probability of B lying = 1 - (Probability of B telling the truth) = 1 - 0.83 = 0.17

Probability of A lying and B telling the truth:
- Probability of A lying = 1 - (Probability of A telling the truth) = 1 - 0.75 = 0.25
- Probability of B telling the truth = 0.83

Now, we can calculate the probability of the two scenarios occurring:
- Probability of scenario 1 = Probability of A telling the truth * Probability of B lying = 0.75 * 0.17 = 0.1275
- Probability of scenario 2 = Probability of A lying * Probability of B telling the truth = 0.25 * 0.83 = 0.2075

Finally, we add the probabilities of the two scenarios to get the total probability of A and B contradicting each other:
- Probability of A and B contradicting each other = Probability of scenario 1 + Probability of scenario 2 = 0.1275 + 0.2075 = 0.335

Therefore, the probability that A and B will contradict each other in stating the same fact is approximately 0.335, which can be simplified to 1/3. Hence, the correct answer is option B.

Similar Quant Doubts

A speaks the truth in 75% and B in 80% cases. In what percentage of cases are they likely to contradict each other when narrating the same incident?

A speak truth in 60% cases and B in 80% cases. In what percent of cases they likely to contradict each other narrating the same incident?

A, B ans C are three speakers. They have to speak randomly along with another 5 speakers in a function.A has to speak before B and B has to speak before C. What is the probability. Correct answer is '1/6'. Can you explain this answer?

If 6 times the no. of permutations of n items taken 3 at a time is equal to 7 times the no. of permutations of (n-1) items taken 3 at a time then the value of n will be?

Top Courses for Quant

Suggested Free Tests

Related Content

Quant Courses

Related Exams

Related Question Answers

Related Searches

Company