Question

50 people, 20 always tell truth.Two selected. Find probability distribution of number of truth-tellers.

Answer 1

Let: \( X \) = Number of truth-tellers selected when 2 people are chosen at random from a group.
Suppose there are 50 people: 20 truth-tellers and 30 liars.
Possible values of \( X \): 0, 1, 2
Case 1: \( X = 0 \) (Both are liars)
\[ P(X = 0) = \frac{30}{50} \cdot \frac{29}{49} = \frac{870}{2450} = \frac{87}{245} \] Case 2: \( X = 2 \) (Both are truth-tellers)
\[ P(X = 2) = \frac{20}{50} \cdot \frac{19}{49} = \frac{380}{2450} = \frac{38}{245} \] Case 3: \( X = 1 \) (One truth-teller, one liar)
Two ways: Truth then lie OR Lie then truth \[ P(X = 1) = \frac{20}{50} \cdot \frac{30}{49} + \frac{30}{50} \cdot \frac{20}{49} = \frac{600 + 600}{2450} = \frac{1200}{2450} = \frac{120}{245} \] 
Therefore, the probability distribution is:
 

X	P(X)
0	\( \frac{87}{245} \)
1	\( \frac{120}{245} \)
2	\( \frac{38}{245} \)