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Dmitrij [34]
3 years ago
12

Suppose that the new england colonials baseball team is equally likely to win a game as not to win it. if 4 colonials games are

chosen at random, what is the probability that exactly 2 of those games are won by the colonials

Mathematics
1 answer:
seraphim [82]3 years ago
7 0
1. Consider the tree diagram in the picture attached.

2 Each of the games picked can be a W (win) or a L (lose), the chances are equal

3. All that could happen in 4 picked games is shown in the picture:

2 w 2 l can happen in the following 6 ways: {(W,W,L,L)(W,L,W,L)(W,L,L,W)(L,W,W,L)(L,W,L,W)(L,L,W,W)}

4. P(2W,2L)=6/16=3/8

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Answer:

E[W] = $25 (assuming the currency is in dollars)

Var(W) = 1041.67

Step-by-step explanation:

Probability of winning first starts with the coin toss.

For a win, the coin needs to land on heads.

Probability of that = 1/2 = 0.5

Then probability of winning any amount = 1/100 = 0.01

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Summing from 0 to 100 (0 indicating getting a tail from the coin toss). This could be done with dome faster with an integral sign

E(W) = ∫ 0.005 x dx

Integrating from 0 to 100

E(W) = [0.005 x²/2]¹⁰⁰₀

E(W) = [0.0025 x²]¹⁰⁰₀ = 0.0025(100² - 0²) = 0.0025 × 10000 = $25

Variance is given by

Variance = Var(X) = Σxᵢ²pᵢ − μ²

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Σxᵢ²pᵢ = Σ 0.005xᵢ²

Σ 0.005 xᵢ² = ∫ 0.005 x² dx

Integrating from 0 to 100

∫ 0.005 x² dx = [0.005 x³/3]¹⁰⁰₀ = [0.1667x³]¹⁰⁰₀ = 0.1667(100³ - 0³) = 1666.67

Var(W) = 1666.67 - 25² = 1666.67 - 625 = 1041.67.

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