Question

Team A plays with $5$ other teams exactly once. Assuming that for each match the probabilities of a win, draw and loss are equal, then

• A. The probability that A wins and loses equal number of matches is $\frac{34}{81}$
• B. The probability that A wins and loses equal number of matches is $\frac{17}{81}$
• C. The probability that A wins more number of matches than it loses is $\frac{17}{81}$
• D. The probability that A loses more number of matches than it wins is $\frac{16}{81}$

Answer 1

The correct option is B The probability that A wins and loses equal number of matches is $\frac{17}{81}$

Probability of equal number of Draw (D), win(W) and loss (L).

i.e., $P\left(D\right)=P\left(W\right)=P\left(L\right)=\frac{1}{3}$

Now the probability of win and loss of equal number of times as following cases.

Case 1:

Win $=0$ and Loss $=0$

Here, all $5$ matches becomes draw.

$\Rightarrow P\left(0W\right).P\left(0L\right)=(\frac{1}{3}{)}^5$ ---(1)

Case 2:

Win $=1$ and Loss $=1$

$\Rightarrow P\left(1W\right).P\left(1L\right){=}^5{C}_1{.}^4{C}_1.(\frac{1}{3}{)}^5$ ---(2)

Case 3:

Win $=2$ and Loss $=2$

$\Rightarrow P\left(2W\right).P\left(2L\right){=}^5{C}_2{.}^3{C}_2.(\frac{1}{3}{)}^5$----(3)

From Case 1, Case 2 and Case 3,

probability of win and loss of equal number of times

$=(\frac{1}{3}{)}^5{+}^5{C}_1{.}^4{C}_1.(\frac{1}{3}{)}^5{+}^5{C}_2{.}^3{C}_2.(\frac{1}{3}{)}^5$

$=\left(\frac{1}{3}{)}^5\right(1{+}^5{C}_1{.}^4{C}_1{+}^5{C}_2{.}^3{C}_2)$

$=\left(\frac{1}{3}{)}^5\right(1+5\left(4\right)+(\frac{5\times 4}{1\times 2}\times \frac{3\times 2}{1\times 2})$

$=\frac{1}{243}(1+20+30)$

$=\frac{51}{243}$

$=\frac{17}{81}$

Hence, Option C is correct.