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Suppose that two teams play a series of games that end when one of them has won i games. Suppose that each game played is, independently, won by team A with probability p. Find the expected number of games that are played when i = 2. Also show that this number is maximized when p= 21.

Sagot :

Batolisisidn

Answer:

a) E(x) = -2p^2 + 2p + 2

b) Number is maximized when p = 1/2

Step-by-step explanation:

Determine the Expected number of games  when ( i ) = 2

The number of possible combinations that both teams win two games :

AA, BB, ABB, ABA, BAA, BAB  = 6 combinations

P( team A winning ) = p

P( team B wins ) = 1 - p

Attached below is the detailed solution on the expected number of games

expected number of games ; E(x) = -2p^2 + 2p + 2

ii) Number is maximized when p = 1/2

View image Batolisis
Lhmarianateixeiraidn

In this exercise we will use the knowledge of probability and combination, so we have what will be:

a)[tex]E(x) = -2p^2 + 2p + 2[/tex]

b)[tex]p = 1/2[/tex]

Organizing the information given in the statement as:

  • Expected number of games  when ( i ) = 2

A)The number of possible combinations that both teams win two games :

[tex]AA, BB, ABB, ABA, BAA, BAB = 6 \ combinations\\P( team\ A \ winning ) = p\\P( team \ B \ wins ) = 1 - p\\E(x) = -2p^2 + 2p + 2[/tex]

B) To calculate the maximum number we must solve the quadratic equation, like this:

[tex]p=1/2[/tex]

See more about probability at brainly.com/question/795909

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