Suppose that two teams play a series of games that ends when one of them has won games. Also suppose that each game played is won by team with probability p, independent of all other games. Find the expected number of games that are played when

(a) E(X) = -2p² + 2p + 2; d²/dp² E(X) at p = 1/2 is less than 0

(b) 6p⁴ - 12p³ + 3p² + 3p + 3; d²/dp² E(X) at p = 1/2 is less than 0

Step-by-step explanation:

(a) when i = 2, the expected number of played games will be:

E(X) = 2[p² + (1-p)²] + 3[2p² (1-p) + 2p(1-p)²] = 2[p²+1-2p+p²] + 3[2p²-2p³+2p(1-2p+p²)] = 2[2p²-2p+1] + 3[2p² - 2p³+2p-4p²+2p³] =  4p²-4p+2-6p²+6p = -2p²+2p+2.

If p = 1/2, then:

d²/dp² E(X) = d/dp (-4p + 2) = -4 which is less than 0. Therefore, the E(X) is maximized.

(b) when i = 3;

E(X) = 3[p³ + (1-p)³] + 4[3p³(1-p) + 3p(1-p)³] + 5[6p³(1-p)² + 6p²(1-p)³]

Simplification and rearrangement lead to:

E(X) = 6p⁴-12p³+3p²+3p+3

if p = 1/2, then:

d²/dp² E(X) at p = 1/2 = d/dp (24p³-36p²+6p+3) = 72p²-72p+6 = 72(1/2)² - 72(1/2) +6 = 18 - 36 +8 = -10

Therefore, E(X) is maximized.

b

2

−

b

b

3

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7

b

2

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14

b

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6

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3

b

3

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7

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2

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14

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6

%

step-by-step explanation: