Question
A speaks truth in 60% cases B speaks truth in 70% cases. The probability that they will way say the same thing while describing a single event, is-
Answer: Option A
Let $${E_1}$$ = event that A speaks the truth
And $${E_2}$$ = event that B speaks the truth
Then,
$$\eqalign{
& P\left( {{E_1}} \right) = \frac{{60}}{{100}} = \frac{3}{5}, \cr
& P\left( {{E_2}} \right) = \frac{{70}}{{100}} = \frac{7}{{10}}, \cr
& P\left( {{{\bar E}_1}} \right) = \left( {1 - \frac{3}{5}} \right) = \frac{2}{5}, \cr
& P\left( {{{\bar E}_2}} \right) = \left( {1 - \frac{7}{{10}}} \right) = \frac{3}{{10}} \cr} $$
P (A and B say the same thing) = P [(A speaks the truth and B speaks the truth) or (A tells a lie and B tells a lie)]
$$=$$ P [$$\left( {{E_1} \cap {E_2}} \right)$$ or $$({\overline E _1} \cap {\overline E _2})] $$
$$=$$ P $$\left( {{E_1} \cap {E_2}} \right)$$ + $$({\overline E _1} \cap {\overline E _2})$$
$$=$$ P ($${{E_1}}$$). P ($${{E_2}}$$) + P ($${\overline E _1}$$) . P ($${\overline E _2}$$)
$$\eqalign{
& = \left( {\frac{3}{5} \times \frac{7}{{10}}} \right) + \left( {\frac{2}{5} \times \frac{3}{{10}}} \right) \cr
& = \frac{{27}}{{50}} \cr
& = 0.54 \cr} $$
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Let $${E_1}$$ = event that A speaks the truth
And $${E_2}$$ = event that B speaks the truth
Then,
$$\eqalign{
& P\left( {{E_1}} \right) = \frac{{60}}{{100}} = \frac{3}{5}, \cr
& P\left( {{E_2}} \right) = \frac{{70}}{{100}} = \frac{7}{{10}}, \cr
& P\left( {{{\bar E}_1}} \right) = \left( {1 - \frac{3}{5}} \right) = \frac{2}{5}, \cr
& P\left( {{{\bar E}_2}} \right) = \left( {1 - \frac{7}{{10}}} \right) = \frac{3}{{10}} \cr} $$
P (A and B say the same thing) = P [(A speaks the truth and B speaks the truth) or (A tells a lie and B tells a lie)]
$$=$$ P [$$\left( {{E_1} \cap {E_2}} \right)$$ or $$({\overline E _1} \cap {\overline E _2})] $$
$$=$$ P $$\left( {{E_1} \cap {E_2}} \right)$$ + $$({\overline E _1} \cap {\overline E _2})$$
$$=$$ P ($${{E_1}}$$). P ($${{E_2}}$$) + P ($${\overline E _1}$$) . P ($${\overline E _2}$$)
$$\eqalign{
& = \left( {\frac{3}{5} \times \frac{7}{{10}}} \right) + \left( {\frac{2}{5} \times \frac{3}{{10}}} \right) \cr
& = \frac{{27}}{{50}} \cr
& = 0.54 \cr} $$
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