In a tournament, team X plays with each of the 6 other teams once. For each match the
probability of a win, draw and loss are equal. If the probability that the team X, finishes with more wins than losses is $\frac{p}{q}$ where p and q are co-prime then the sum of digits of $| p - q |$ is.

Question

Infinity Learn · Accepted Answer

The correct option is B 294
Total number of results

= 3^{6} = 729

Required number of ways

= \frac{1}{2} (729 - (\begin{array}{l} N u m b e r o f w a y s i n w h i c h \\ t e a m T f i n i s h e w i t h e q u a l n u m b e r \\ o f w i n s a n d l o s s e s \end{array}))

Now, we shall consider following cases:
Case 1: 0 draw, 3 wins and 3 losses
WWWLLL
Number of ways

= \frac{6!}{3! 3!} = {}^{6}C_{3} = 20

Case II: 1 win, 1loss, 4draws:
WWLLDD
Number of ways

= \frac{6!}{4!} = 30

Case III: 2 wins, 2losses, 2draws: WWLLDD
Number of ways

P (A) = \frac{999}{1000} \times \frac{998}{999} \times .......... \times \frac{1000 - k + 1}{1000 - k + 2} \times \frac{1}{1000 - k + 1}

Case IC: no win and no loss, 6draws: DDDDDD
Number of ways=1
Total ways with equal wins and draws

= 30 + 30 + 90 + 4 = 141

So, required number of ways

= \frac{1}{2} (729 - 141) = \frac{588}{2} = 294

Detailed Solution