P speaks truth in 70 % of the cases and Q in 80 % of the cases. In what percent of cases are they likely to…

Let $A$ and $B$ denote the events $P$ speaks the truth and $Q$ speaks the truth respectively.

Given $P\left(A\right)=\frac{70}{100}$ and $P\left(B\right)=\frac{80}{100}$

Hence, $P\left(\overline{A}\right)=1-P\left(A\right)$ and $P\left(\overline{B}\right)=1-P\left(B\right)$

$\Rightarrow P\left(\overline{A}\right)=1-\frac{70}{100}=\frac{30}{100}$ and $P\left(\overline{B}\right)=1-\frac{80}{100}=\frac{20}{100}.$

We have to find the probability that they agree in stating the same fact this means either both of them are stating it correctly or both of them are stating it wrongly.

The required probability is $=P\left(A\right)P\left(B\right)+P\left(\overline{A}\right)P\left(\overline{B}\right)$

$=\left(\frac{70}{100}\right)\left(\frac{80}{100}\right)+\left(\frac{30}{100}\right)\left(\frac{20}{100}\right)$

$=\frac{56}{100}+\frac{6}{100}$

$=\frac{62}{100}=62\%.$

Therefore, the percentage of cases in which they are likely to agree is $62\%$.