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\mathrm{A } speaks truth in 60 \% cases and \mathrm{B} speaks truth in 70 \% cases. The probability that they will not say the same thing while describing a single event is
 

Option: 1

0.54
 


Option: 2

0.46


Option: 3

0.38


Option: 4

0.94


Answers (1)

best_answer

Consider the following events

\mathrm{A_1 ; A} speaks truth

\mathrm{A_2: B} speaks truth

Then, \mathrm{P\left(A_1\right)=\frac{60}{100}=\frac{3}{5}, P\left(A_2\right)=\frac{70}{100}=\frac{7}{10}}

For the required event, either of them should speak a truth.

Required probability

\mathrm{ =P\left[\left(A_1 \cap \overline{A_2}\right) \cup\left(\overline{A_1} \cap A_2\right)\right] }

\mathrm{ =P\left(A_1 \cap \bar{A}_2\right)+P\left(\bar{A}_1 \cap A_2\right) }

\mathrm{ =P\left(A_1\right) P\left(\bar{A}_2\right)+P\left(\bar{A}_1\right) P\left(A_2\right) }

\mathrm{=\frac{3}{5} \times\left(1-\frac{7}{10}\right)+\left(1-\frac{3}{5}\right) \times \frac{7}{10} }

\mathrm{ =\frac{3}{5} \times \frac{3}{10}+\frac{2}{5} \times \frac{7}{10}=0.46}

Hence option 2 is correct.







 

Posted by

shivangi.shekhar

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