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Worked solution

Separate dependent sets from independent repeated matches

A and B play a racquet match of at most 3 sets. The first to win two sets wins. A wins the first set with probability 0.6. After the first set, A wins the next set with probability p given A won the preceding set; B wins the next set with probability 0.7 given B won the preceding set.

Find, in terms of p, the probability that A wins the match.

(i) The probability B won the first set given B lost the match is 0.15. Find p. marks

(ii) Given instead p=0.65 and that A’s third win occurs in the nth match with probability 0.10866 correct to 5 significant figures, write an equation for n and solve it numerically. marks

[2014/NYJC/II/6]

CJC · Review problems · Question 12

Answer

P(A wins match)=0.18+0.54pP(A\text{ wins match})=0.18+0.54p. (i) p=9/13≈0.692p=9/13\approx0.692; (ii) n=7n=7.

Retain the source’s stopping tree and winning paths

The source uses the previous-set transition model, so after B the next-A probability is 0.3. A wins on AA, ABA, BAA, with probabilities 0.6p0.6p, 0.6(1−p)(0.3)0.6(1-p)(0.3) and 0.4(0.3)p0.4(0.3)p. Their sum is 0.6p+0.6(1−p)(0.3)+0.4(0.3)p=0.18+0.54p0.6p+0.6(1-p)(0.3)+0.4(0.3)p=0.18+0.54p. No third set is played after AA or BB.

How did this part go?

(i) B-first but A wins means BAA

B losing the match is A winning. Hence

0.15=P(BAA)P(A wins match)=0.4(0.3)p0.18+0.54p.\begin{aligned}&0.15\\ & =\frac{P(BAA)}{P(A\text{ wins match})}\\ & =\frac{0.4(0.3)p}{0.18+0.54p}.\end{aligned}

Thus 0.027+0.081p=0.12p0.027+0.081p=0.12p, so p=9/13≈0.692p=9/13\approx0.692, retaining the source exact/decimal forms.

How did this part go?

(ii) Use the source independent-match model

With p=0.65, A’s match-win probability is q=0.18+0.54(0.65)=0.531q=0.18+0.54(0.65)=0.531. The repeated-match calculation assumes matches are independent with unchanged win probability q. This does not make sets within a match independent; their transitions remain as given.

How did this part go?

(ii) Third match win on trial n

The first n−1 matches must contain exactly two A wins, followed by an A win on match n. The source equation is

(n−12)(0.531)2(1−0.531)n−3(0.531)=0.10866,\binom{n-1}2(0.531)^2(1-0.531)^{n-3}(0.531)=0.10866,

equivalently

(n−1)(n−2)2(0.531)3(0.469)n−3=0.10866.\begin{aligned}&\frac{(n-1)(n-2)}2(0.531)^3(0.469)^{n-3}\\ & =0.10866.\end{aligned}

Numerically, the integer solution is n=7. The supplied probability is rounded: for n=7 the left side is approximately 0.1086591224, which rounds to 0.10866 at 5 s.f. The rounded equality is interpreted accordingly.

How did this part go?

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Initial A/B0.6/0.4. After an A set, next A/Bp,1−p; after a B set, next A/B0.3,0.7. Stop at AA or BB. Continue AB to ABA/ABB and BA to BAA/BAB. All ten source branches and six terminal paths retained.
Initial A/B0.6/0.4. After an A set, next A/Bp,1−p; after a B set, next A/B0.3,0.7. Stop at AA or BB. Continue AB to ABA/ABB and BA to BAA/BAB. All ten source branches and six terminal paths retained.
Initial A/B0.6/0.4. After an A set, next A/Bp,1−p; after a B set, next A/B0.3,0.7. Stop at AA or BB. Continue AB to ABA/ABB and BA to BAA/BAB. All ten source branches and six terminal paths retained.

Common mistakes

Match-level independence in(ii) does not justify treating the sets as independent. Respect both the stopping rule and the rounded target probability.

Solution notes

Initial unnumbered ask, both marked asks, full stopping tree, all source path products, conditional equation, both negative-binomial equation forms and numerical answer retained. Source repeated-match independence is explicit; target 0.10866 is recognised as 5 s.f., not an exact rational constraint.

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