Teams T1, T2, T3 and T4 are in the playoffs. In the semifinal matches T1 plays T4 and T2 plays T 3. The winners of those two matches will play each other in the final match to determine the champion. When Ti plays Tj the probability that Ti wins is (i/i+j) and the outcomes of all the matches are independent. The probability that T 4 will be the champion is ( p / q ), where p and q are relatively prime positive integers. Find the value of (q-13/p).

Question

Teams T1, T2, T3 and T4 are in the playoffs. In the semifinal matches T1 plays T4 and T2 plays T 3. The winners of those two matches will play each other in the final match to determine the champion. When Ti plays Tj the probability that Ti wins is (i/i+j) and the outcomes of all the matches are independent. The probability that T 4 will be the champion is ( p / q ), where p and q are relatively prime positive integers. Find the value of (q-13/p). (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

P(E) =P(P4. and P2 and .P4)+(P4. and P3 and .P4) =(4/5) × (2/5) × (4/6)+(4/5) × (3/5) × (4/7) =(32/150)+(48/175)=(256/525)=( p / q ) ∴ ( q -13/ p )=(512/256)=2

P(E)=P(P4​ and P2​ and P4​)+(P4​ and P3​ and P4​) =54​×52​×64​+54​×53​×74​ =15032​+17548​=525256​=qp​ ∴pq−13​=256512​=2

$P (E) = P (P_{4}$ and $P_{2}$ and $P_{4}) + (P_{4}$ and $P_{3}$ and $P_{4})$
$= \frac{4}{5} \times \frac{2}{5} \times \frac{4}{6} + \frac{4}{5} \times \frac{3}{5} \times \frac{4}{7}$
$= \frac{32}{150} + \frac{48}{175} = \frac{256}{525} = \frac{p}{q}$
$∴ \frac{q - 13}{p} = \frac{512}{256} = 2$