Question

Q4. A speaks truth in 60 percent of the cases and B in 90 percent of the cases. The percentage of cases they are likely to contradict each other in stating the same fact is:

• A. (a) 36 percent
• B. (b) 48 percent
• C. (c) 42 percent
• D. (d) None of the above

Answer 1

The correct option is C

(c) 42 percent

Let E be the event that A speaks the truth and F be the event that B speaks the truth.

E and F are complementary events in which A and B tell a lie.

$Here,P\left(E\right)=\frac{60}{100}=\frac{3}{5},P\left(E\right)=1-P\left(E\right)=\frac{2}{5}P\left(E\right)=\frac{90}{100}=\frac{9}{10},P\left(E\right)=1-P\left(E\right)=\frac{1}{10}$

A and B will contradict each other when one speaks the truth and other tells a lie and this can happen in the following two mutually exclusive ways:

1. A speaks the truth and B tells a lie, i.e., $E\cap \overline{F}$

2. A tells a lie and B speaks the truth, i.e., $\overline{E}\cap F$

Reqd. probability = P $E\cap \overline{F}$ + $\overline{E}\cap F$

=P (E) P $\overline{F}$ + P $\overline{E}$ P (F)

$\frac{3}{5}\times \frac{1}{10}\times \frac{2}{5}\times \frac{9}{10}=\frac{21}{50}$

Hence, they are expected to contradict each other

in $\frac{21}{50}\times 100=42$ %