Teams named East and West are playing a best-2-out-of-3 series of games. This me

Question

Teams named East and West are playing a best-2-out-of-3 series of games. This means that the first team to win two games wins the series. (For example, in the NCAA College Baseball World Series, the two teams in the finals play a best-2-out-of-3 series.) If the first two games are won by the same team, the series is over. The 3rd game is played only if the teams are 1-1 after the first two games.

In each game that is played, the probability that East wins is p. So the probability West wins is q = 1 - p.

Suppose the probability that the winner of the first game ends up winning the series is 0.92.

There are two different values of p for which this can be true.

The smaller value of p which makes this true is  .
The larger value of p which makes this true is  .

Hint: To do this problem you'll need to solve a quadratic equation. So you'll have to use the quadratic formula or a graphing tool such as the Desmos Graphing Calculator on this problem.

Draw a tree diagram for the series of games played, and enter "p" or "q" on each branch of the tree. So for each possible outcome, you'll be able to express the probability of the outcome as a product of p's and q's. Write out the sum of the probabilities of all the outcomes where the winner of the series is the same team that wins wins the first game of the series. (So this will be a sum of terms involving p's and q's. Now replace every "q" by "1-p" so that "p" is the only unknown, and simplify this expression as much as possible by combining terms. You'll wind up with a quadratic expression involving the unknown "p". Then set this expression equal to 0.92 and solve for the two values of p that satisfy this equation. If you're using the quadratic formula, then you'll get both values out of the quadratic formula. If you're using a graphing tool, you'll simply be identifying the two places where a parabola intersects a straight line.

you do not need to draw tree diagram! thanks

Explanation / Answer

First Game -

Case E :  East wins with probability p.

Case W :  West wins with probability q.

Second Game :

Case EE: East win in first game and then in 2nd game with proability = p2

Case EW: East win in first game and West in 2nd game with proability = pq

Case WW: West win in first game and then in 2nd game with proability = q2

Case WE: West win in first game and East in 2nd game with proability = pq

Third Game : (will happen when teams are 1-1 after the first two games)

Case EWE: East win in first game and West in 2nd game and then East win in last game with proability = p2q

Case EWW: East win in first game and West in 2nd game and then West win in last game with proability = pq2

Case WEW: West win in first game and East in 2nd game and then West win in last game with proability = pq2

Case WEE: West win in first game and East in 2nd game and then East win in last game with proability = p2q

Probability that the winner of the first game ends up winning the series = P(Case EE) + P(Case WW) + P(Case EWE) + P(Case WEW) = 0.92

=> p2 + q2 + p2q + pq2 = 0.92

Substituting q = 1-p , we get

=> p2 + (1-p)2 + p2(1-p) + p(1-p)2 = 0.92

=> p2 + 1 + p2 - 2p + p2 - p3 + p(1 + p2 -2p) = 0.92

=> p2 + 1 + p2 - 2p + p2 - p3 + p + p3 -2p2 = 0.92

=> p2 + 1 - p = 0.92

=> p2 - p + 0.08 = 0

Using  quadratic formula, we get

p = [1 - sqrt(12 - 4*1*0.08)] / 2 or  [1 + sqrt(12 - 4*1*0.08)] / 2

p = [1 - sqrt(0.68)] / 2 or [1 + sqrt(0.68)] / 2

p = (1- 0.8246) / 2 or (1+ 0.8246) / 2

p = 0.0877 or 0.9123

So,

The smaller value of p which makes this true is 0.0877 .
The larger value of p which makes this true is  0.9123.

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