Question

Three persons A, B and C planned to have a running race among themselves. If the probability that A wins the race is three times that of B and the probability that B wins the race is \(\dfrac32\) times that of C, then the difference in probabilities of A and C to win the race is

• A. \(\dfrac23\)
• B. \(\dfrac12\)
• C. \(\dfrac5{14}\)
• D. \(\dfrac37\)

Hint:

When probabilities are given in ratios, first express them using a common variable, then use \[ \boxed{P(A)+P(B)+P(C)=1} \] to determine the unknown probability.

Answer 1

The Correct Option is B

Step 1: Assume the probability of C winning is \(x\). Then, \[ P(C)=x. \] Since \[ P(B)=\frac32x, \] and \[ P(A)=3P(B) = 3\left(\frac32x\right) = \frac92x. \]

Step 2: Use the total probability. Since one of the three persons must win, \[ P(A)+P(B)+P(C)=1. \] Therefore, \[ \frac92x+\frac32x+x=1. \] Multiplying by \(2\), \[ 9x+3x+2x=2, \] \[ 14x=2, \] \[ x=\frac17. \] Hence, \[ P(C)=\frac17, \] \[ P(B)=\frac{3}{14}, \] and \[ P(A)=\frac{9}{14}. \]

Step 3: Find the required difference. \[ P(A)-P(C) = \frac9{14}-\frac17 = \frac9{14}-\frac2{14} = \frac7{14} = \boxed{\frac12}. \] Hence, the correct option is \(\boxed{(B)}\).