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A Speaks Truth in 75% and B in 80% of the Cases. in What Percentage of Cases Are They Likely to Contradict Each Other in Narrating the Same Incident? - Mathematics

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Question

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

 
Sum
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Solution

\[P\left( \text{ Both narrating different incident }  \right) = P\left( \text{ Blies and A speaks the truth } \right) + P\left( \text{ A lies and B speaks the truth } \right)\]
\[ = P\left( A \cap \bar{B} \right) + P\left( \bar{A} \cap B \right)\]
\[ = P\left( A \right)P\left( \bar{B} \right) + P \bar{\left( A \right)}P\left( B \right)\]
\[ = 0 . 75\left( 1 - 0 . 8 \right) + \left( 1 - 0 . 75 \right)0 . 8\]
\[ = 0 . 75 \times 0 . 2 + 0 . 25 \times 0 . 8\]
\[ = 0 . 15 + 0 . 2\]
\[ = 0 . 35\]
\[ = 35 \%\]

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Problems based on Probability
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Chapter 31: Probability - Exercise 31.5 [Page 68]

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RD Sharma Mathematics [English] Class 12
Chapter 31 Probability
Exercise 31.5 | Q 5 | Page 68
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