NIMCET 2023 — Mathematics PYQ

1. Convert Percentages into Probabilities

Let $P(A)$ be the probability that A speaks the truth, and $P(B)$ be the probability that B speaks the truth:

$$P(A)=60\%=\frac{60}{100}=\frac{3}{5}$$

$$P(B)=50\%=\frac{50}{100}=\frac{1}{2}$$

Now, find the probabilities that they lie (do not speak the truth), denoted by $P(A^{\prime })$ and $P(B^{\prime })$:

$$P(A^{\prime })=1-P(A)=1-\frac{3}{5}=\frac{2}{5}$$

$$P(B^{\prime })=1-P(B)=1-\frac{1}{2}=\frac{1}{2}$$

2. Condition for Contradiction

Two people contradict each other when one speaks the truth and the other tells a lie. This can happen in two mutually exclusive cases:

1. Case I: A speaks the truth AND B lies $⟹P(A)\cdot P(B^{\prime })$

2. Case II: A lies AND B speaks the truth $⟹P(A^{\prime })\cdot P(B)$

Since these events are independent, the total probability of contradiction $P(C)$ is given by:

$$P(C)=[P(A)\cdot P(B^{\prime })]+[P(A^{\prime })\cdot P(B)]$$

3. Substitute Values and Calculate

$$P(C)=\left(\frac{3}{5}\cdot \frac{1}{2}\right)+\left(\frac{2}{5}\cdot \frac{1}{2}\right)$$

$$P(C)=\frac{3}{10}+\frac{2}{10}$$

$$P(C)=\frac{5}{10}=\frac{1}{2}$$