$P_{1},P_{2},P_{3}$ and $P_{4}$ are four players playing in a knockout tournament and probability th

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$P_{1},P_{2},P_{3}$ and $P_{4}$ are four players playing in a knockout tournament and probability that $P_{1}$ will pair up with $P_{i}$ is proportional to i and $P_{m}$ wins with $P_{n}$ if $m < n$; then the probability that $P_{2}$ will reach the second round is

Accepted Answer

Sol. [Key Concept: $P_{1}..P_{4}$ knockout with random pairing: probability of the stated event.]$P_{1}$ get paired with $P_{2}$ in $1^{	ext{st}	ext{~}}$ round $= 2k$where $K(2 + 3 + 4) = 1\ \Rightarrow \ k = \frac{1}{9}$$	herefore\ P\left( P_{2} ight.\ $ reaches second round $) = 1 - P\left( P_{2} ight.\ $ paired with $\left. \ P_{1} ight)$$$= 1 - \frac{2}{9} = \frac{7}{9}$$

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