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Team A and team B play a series. The first team to win three games wins the series. Each team is equally likely to win each game, there are no ties, and the outcomes of the individual games are independent. If team B wins the second game and team A wins the series, what is the probability that team B wins the first game?

Answer :

Answer:

Probability is [tex]\frac{1}{4}[/tex]

Step-by-step explanation:

This problem can be easily done tree diagram.

Few things keep in mind :

  • From starting of match three victories of A.
  • Less then or equal to two victories of B.
  • In each matches there are two possibilities either favourable for A or B.

First case if A wins first match

  1.       two victories of B then two of A
  2.       victory of B then A then B then A
  3.       victory of B then A then A

Second case if B wins first match(favorable)

  1.      victory of two B then three A

So total cases are 4 and favorable is one

  probability=[tex]\frac{1}{4}[/tex]

[Tree diagram is in attachment]

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