Worksheet on Expected Value - Free Printable

Let's solve each problem step by step, constructing the probability distribution and calculating the expected value for each scenario.

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1) Raffle Tickets

- 2000 tickets sold at $2 each
- One prize: $1500 computer system
- A person buys one ticket

#### Step 1: Determine possible outcomes
- Win: $1500 (net gain = $1500 - $2 = $1498)
- Lose: $0 (net gain = -$2)

#### Probabilities:
- P(win) = 1/2000
- P(lose) = 1999/2000

#### Expected Value (EV):
$$
EV = \left(\frac{1}{2000} \times 1498\right) + \left(\frac{1999}{2000} \times (-2)\right)
$$
$$
= \frac{1498}{2000} - \frac{3998}{2000} = \frac{-2500}{2000} = -1.25
$$

✔ Expected Value = -$1.25

> This means on average, a person loses $1.25 per ticket.

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2) Ski Resort Tickets

- 1000 tickets sold at $1 each
- Prizes:
- $150 (1 prize)
- $75 (1 prize)
- $25 (1 prize)
- $5 (1 prize)
- Person buys two tickets

We need expected value for one person who buys two tickets.

But since tickets are randomly drawn, we assume without replacement, but with large number of tickets, we can approximate using probabilities.

However, let’s compute expected value per ticket, then double it.

#### Step 1: Find expected value per ticket

Each ticket costs $1.

Possible prizes:
- $150 → prob = 1/1000
- $75 → prob = 1/1000
- $25 → prob = 1/1000
- $5 → prob = 1/1000
- $0 → prob = 996/1000

Net gains (prize minus cost):
- Win $150 → net gain = $149
- Win $75 → net gain = $74
- Win $25 → net gain = $24
- Win $5 → net gain = $4
- Lose → net gain = -$1

#### EV per ticket:
$$
EV = \left(\frac{1}{1000} \times 149\right) + \left(\frac{1}{1000} \times 74\right) + \left(\frac{1}{1000} \times 24\right) + \left(\frac{1}{1000} \times 4\right) + \left(\frac{996}{1000} \times (-1)\right)
$$
$$
= \frac{149 + 74 + 24 + 4}{1000} - \frac{996}{1000}
= \frac{251}{1000} - \frac{996}{1000} = \frac{-745}{1000} = -0.745
$$

So EV per ticket = -$0.745

For two tickets:
$$
EV = 2 \times (-0.745) = -1.49
$$

✔ Expected Value = -$1.49

> On average, someone loses $1.49 when buying two tickets.

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3) Box with Bills

- 10 × $1 bills
- 5 × $2 bills
- 3 × $5 bills
- 1 × $10 bill
- 1 × $100 bill

Total bills = 10 + 5 + 3 + 1 + 1 = 20 bills

Person pays $20 to select one bill.

We want expected value of the amount received, then subtract cost to get net expected value.

#### Step 1: Compute expected value of the bill selected

| Bill | Amount | Count | Probability |
|------|--------|-------|-------------|
| $1 | 1 | 10 | 10/20 = 0.5 |
| $2 | 2 | 5 | 5/20 = 0.25 |
| $5 | 5 | 3 | 3/20 = 0.15 |
| $10 | 10 | 1 | 1/20 = 0.05 |
| $100 | 100 | 1 | 1/20 = 0.05 |

$$
EV_{\text{bill}} = (1)(0.5) + (2)(0.25) + (5)(0.15) + (10)(0.05) + (100)(0.05)
$$
$$
= 0.5 + 0.5 + 0.75 + 0.5 + 5 = 7.25
$$

So expected value of the bill is $7.25

But person paid $20, so net expected value:

$$
EV = 7.25 - 20 = -12.75
$$

✔ Expected Value = -$12.75

> On average, the person loses $12.75.

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4) SADD Cash Prize

- $100 cash prize
- 875 tickets sold at $2 each
- One person buys one ticket

#### Outcomes:
- Win: $100 → net gain = $100 - $2 = $98
- Lose: $0 → net gain = -$2

P(win) = 1/875
P(lose) = 874/875

$$
EV = \left(\frac{1}{875} \times 98\right) + \left(\frac{874}{875} \times (-2)\right)
$$
$$
= \frac{98}{875} - \frac{1748}{875} = \frac{-1650}{875} = -1.8857...
$$

Rounded to nearest cent: -$1.89

✔ Expected Value = -$1.89

> On average, a person loses $1.89 per ticket.

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5) Ski Resort Profit Expectation

- Losses: $85,000 if not enough snow (bad season)
- Profit: $300,000 if good season (≥75 inches)
- P(good season) = 45% = 0.45
- So P(bad season) = 1 - 0.45 = 0.55

Let’s define profit as:
- Good season: +$300,000
- Bad season: -$85,000

$$
EV = (0.45)(300,000) + (0.55)(-85,000)
$$
$$
= 135,000 - 46,750 = 88,250
$$

✔ Expected Value = $88,250

> The ski resort expects to make an average profit of $88,250 per season.

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✔ Final Answers:

1) Expected Value = -$1.25
2) Expected Value = -$1.49
3) Expected Value = -$12.75
4) Expected Value = -$1.89
5) Expected Value = $88,250

Let me know if you'd like these formatted in a table or with diagrams!