+1 vote
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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.

edited | 211 views

+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
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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
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