G: Consider the following game. A coin is flipped. If it lands tails we stop playing and winnothing. If it lands heads we spin a wheel with numbers from1to100, each with equalprobability. We get the amount of money written on the weel and we play again (i. e. weflip the coin again). LetWbe our total winnings from all tosses of the coin. FindE[W]andVar(W).

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

Total probability of winning any amount = 0.5 × 0.01 = 0.005

But expected value is given by

E(X) = Σ xᵢpᵢ

where xᵢ is each amount that could be won

pᵢ is the probability of each amount to be won and it is the same for all the possible winnings = 0.005

So,

E(W) = Σ 0.005 xᵢ

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

μ = expected value

We calculate the expression, Σxᵢ²pᵢ which is another sum from 0 to 100

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.

yes because they have the same amount of green and blur marbles.

step-by-step explanation: