
A speaks the truth in 60% of the cases, and B in 90% of the cases. In what percentage of cases are they likely to state the same fact?
Answer
596.1k+ views
Hint: Probability ( an event ) $ = \dfrac{{{\rm{\text{number of favourable outcomes}}}}}{{{\rm{\text{total number of outcomes}}}}}$
Same fact would be given by them only if both speak the truth or both do not speak the truth.
Percentage could be converted into probability and probability has a total value of 1.
If probability of occurring of something is p then probability of not occurring would be 1-p.
Complete step by step solution:
Step 1
Calculate the probability of A and B speaking truth
\[\begin{array}{l}
{\rm{\text{Probability of A speaking truth = }}}{{\rm{P}}_A} = \dfrac{{60}}{{100}} = \dfrac{6}{{10}}\\
{\rm{\text{Probability of B speaking truth =} }}{{\rm{P}}_B} = \dfrac{{90}}{{100}} = \dfrac{9}{{10}}
\end{array}\]
Here, $P_A$ and $P_B$ come out to be 6/10 and 9/10.
Step 2
Calculate the probability of A and B not speaking truth\[\begin{array}{l}
{\rm{\text{Probability of A not speaking truth = }}}{{\rm{Q}}_{{\rm{A}}'}} = 1 - {{\rm{P}}_A} = 1 - \dfrac{6}{{10}} = \dfrac{4}{{10}}\\
{\rm{\text{Probability of B not speaking truth = }}}{{\rm{Q}}_{{\rm{B}}'}} = 1 - {{\rm{P}}_B} = 1 - \dfrac{9}{{10}} = \dfrac{1}{{10}}
\end{array}\]
Step 3
They are likely to provide the same fact when both speaking truth and both not speaking truth.
Step 4
$\begin{array}{l}
\therefore {\rm{\text{Probability(both states the same fact) = Probability(both A and B tell the truth)}}}\\
{\rm{ + }}\\
{\rm{\text{Probability(both A and B lie)}}}
\end{array}$
\[{{\rm{P}}_{\rm{s}}}{\rm{ = }}{{\rm{P}}_{\rm{A}}}{\rm{ \times }}{{\rm{P}}_{\rm{B}}}{\rm{ + }}{{\rm{Q}}_{\rm{A}}}{\rm{ \times }}{{\rm{Q}}_B}\]
Substitute the values and find
\[\begin{array}{l}
{{\rm{P}}_{\rm{S}}} = \dfrac{6}{{10}} \times \dfrac{9}{{10}} + {\rm{ }}\dfrac{4}{{10}} \times \dfrac{1}{{10}}\\
{{\rm{P}}_{\rm{S}}} = {\rm{ }}\dfrac{{54}}{{100}} + \dfrac{4}{{100}} = \dfrac{{58}}{{100}}
\end{array}\]
Here, PS is the probability of stating the same fact which comes out to be 58/100.
Step 5
Express the probability in percentage
\[{\rm{\text{Percentage of}}}\dfrac{{58}}{{100}} = \dfrac{{58}}{{100}} \times 100 = 58\% \]
The percentage comes out to be 58%.
Therefore, In 58% of the cases A and B, are likely to state the same fact.
Note:
Probability of any event P(A), is always “greater than and equal to 0” and “less than and equal to 1”.
$0 \le {\rm{P(A)}} \le {\rm{1}}$.
‘AND’ operations are substituted with $' \times '$, implying both are likely to happen together.
‘OR’ operations are substituted with $' + '$, implying either of the one possibility is likely to happen.
Same fact would be given by them only if both speak the truth or both do not speak the truth.
Percentage could be converted into probability and probability has a total value of 1.
If probability of occurring of something is p then probability of not occurring would be 1-p.
Complete step by step solution:
Step 1
Calculate the probability of A and B speaking truth
\[\begin{array}{l}
{\rm{\text{Probability of A speaking truth = }}}{{\rm{P}}_A} = \dfrac{{60}}{{100}} = \dfrac{6}{{10}}\\
{\rm{\text{Probability of B speaking truth =} }}{{\rm{P}}_B} = \dfrac{{90}}{{100}} = \dfrac{9}{{10}}
\end{array}\]
Here, $P_A$ and $P_B$ come out to be 6/10 and 9/10.
Step 2
Calculate the probability of A and B not speaking truth\[\begin{array}{l}
{\rm{\text{Probability of A not speaking truth = }}}{{\rm{Q}}_{{\rm{A}}'}} = 1 - {{\rm{P}}_A} = 1 - \dfrac{6}{{10}} = \dfrac{4}{{10}}\\
{\rm{\text{Probability of B not speaking truth = }}}{{\rm{Q}}_{{\rm{B}}'}} = 1 - {{\rm{P}}_B} = 1 - \dfrac{9}{{10}} = \dfrac{1}{{10}}
\end{array}\]
Step 3
They are likely to provide the same fact when both speaking truth and both not speaking truth.
Step 4
$\begin{array}{l}
\therefore {\rm{\text{Probability(both states the same fact) = Probability(both A and B tell the truth)}}}\\
{\rm{ + }}\\
{\rm{\text{Probability(both A and B lie)}}}
\end{array}$
\[{{\rm{P}}_{\rm{s}}}{\rm{ = }}{{\rm{P}}_{\rm{A}}}{\rm{ \times }}{{\rm{P}}_{\rm{B}}}{\rm{ + }}{{\rm{Q}}_{\rm{A}}}{\rm{ \times }}{{\rm{Q}}_B}\]
Substitute the values and find
\[\begin{array}{l}
{{\rm{P}}_{\rm{S}}} = \dfrac{6}{{10}} \times \dfrac{9}{{10}} + {\rm{ }}\dfrac{4}{{10}} \times \dfrac{1}{{10}}\\
{{\rm{P}}_{\rm{S}}} = {\rm{ }}\dfrac{{54}}{{100}} + \dfrac{4}{{100}} = \dfrac{{58}}{{100}}
\end{array}\]
Here, PS is the probability of stating the same fact which comes out to be 58/100.
Step 5
Express the probability in percentage
\[{\rm{\text{Percentage of}}}\dfrac{{58}}{{100}} = \dfrac{{58}}{{100}} \times 100 = 58\% \]
The percentage comes out to be 58%.
Therefore, In 58% of the cases A and B, are likely to state the same fact.
Note:
Probability of any event P(A), is always “greater than and equal to 0” and “less than and equal to 1”.
$0 \le {\rm{P(A)}} \le {\rm{1}}$.
‘AND’ operations are substituted with $' \times '$, implying both are likely to happen together.
‘OR’ operations are substituted with $' + '$, implying either of the one possibility is likely to happen.
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