Question:
A person P speaks truth in 75% cases and another person R in 80% cases. Then the probability that they likely to contradict each other in narrating the same event is
A person P speaks truth in 75% cases and another person R in 80% cases. Then the probability that they likely to contradict each other in narrating the same event is
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Two people contradict each other if one tells the truth and the other tells a lie.
P(A contradicts B) = P(A true, B false) + P(A false, B true).
If A and B are independent, P(A true, B false) = P(A true) * P(B false).
Updated On: May 26, 2025
- \( \frac{7}{20} \)
- \( \frac{7}{10} \)
- 0.2
- 0.3
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The Correct Option is A
Solution and Explanation
Let P be the event that person P speaks the truth. \(P(P) = 75% = 0.75 = 3/4\).
Let P' be the event that person P lies. \(P(P') = 1 - 0.75 = 0.25 = 1/4\).
Let R be the event that person R speaks the truth. \(P(R) = 80% = 0.80 = 4/5\).
Let R' be the event that person R lies. \(P(R') = 1 - 0.80 = 0.20 = 1/5\).
They contradict each other if one speaks the truth and the other lies.
There are two mutually exclusive cases for contradiction:
Case 1: P speaks truth AND R lies. Probability = \(P(P \cap R')\).
Case 2: P lies AND R speaks truth. Probability = \(P(P' \cap R)\).
Assuming their statements are independent:
\(P(P \cap R') = P(P) \times P(R') = (3/4) \times (1/5) = 3/20\).
\(P(P' \cap R) = P(P') \times P(R) = (1/4) \times (4/5) = 4/20\).
The probability that they contradict each other is the sum of probabilities of these two cases:
\(P(\text{contradict}) = P(P \cap R') + P(P' \cap R) = 3/20 + 4/20 = 7/20\).
As a decimal, \(7/20 = 35/100 = 0.35\).
Option (a) is \(7/20\).
Option (c) 0.2 = 4/20. Option (d) 0.3 = 6/20.
So, the answer is \(7/20\).
\[ \boxed{\frac{7}{20}} \]
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