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You have to play three games with opponents A and B in a specified sequence. You win the series if you win two consecutive games. A is a stronger player than B. Which sequence maximizes your chance of winning the series?

  1. AAB
  2. ABA
  3. BAB
  4. BAA
  5. All are the same.
asked in Quantitative Aptitude by (560 points) 3 9
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2 Answers

+1 vote
A is a stronger player than B so probability of wining A is higher then probability of wining B .

let  W=win L=lost

Probability of wining A to you = 3/4 , W= 0.75 L=0.25

Probability of wining B to you = 1/2=0.50 L=0.5

Case 1: AAB     

probability for you to win series = WWL + LWW+WWW= 0.25*0.25*0.5+0.75*0.25*0.5+0.25*0.25*0.5=0.15625

Case 2: ABA

probability for you to win series = WWL + LWW+WWW= 0.25*0.5*0.75+0.75*0.5*0.25+0.25*0.5*0.25=0.21875

Case 3: BAB

probability for you to win series = WWL + LWW+WWW=0.5*0.25*0.5+0.5*0.25*0.5+0.5*0.25*0.5=0.1875

Case 4: BAA

probability for you to win series = WWL + LWW+WWW=0.15625

so ABA sequence will maximize our chances to win seriese Ans=2
answered by (270 points) 1 1 7
0

i thank answer is B.

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0 votes
Option B is correct because You must win 2nd game to will series, So ABA  not other bcz  A is a stronger player than B
answered by (560 points) 3 9

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