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Expected Value & Binomial Probability — Easy Worksheet 1
Grade 12 · Whoobee Name: ____________________ Date: __________
Solve each problem. Give probabilities as fractions in simplest form or as decimals rounded the way the problem says.
How to solve
Expected value: E(X) = Σ x·P(x), multiply each outcome by its probability and add. Binomial: with n independent trials and success probability p, P(X = k) = C(n, k)·pk·(1 - p)n-k, and the mean is np.
Example
The table shows the probability distribution of X. Find the expected value E(X).
| x | 1 | 4 | 8 |
|---|---|---|---|
| P(x) | 0.3 | 0.2 | 0.5 |
Answer: 5.1
| x | 2 | 7 | 8 |
|---|---|---|---|
| P(x) | 0.3 | 0.1 | 0.6 |
Not quite. The answer is
💡 E(X) = Σ x·P(x): multiply each value by its probability and add. E(X) = 2 × 0.3 + 7 × 0.1 + 8 × 0.6 = 0.6 + 0.7 + 4.8 = 6.1.
✓ Correct!
Not quite. The answer is
💡 E(X) = Σ x·P(x): multiply each value by its probability and add. E(X) = 2 × 0.3 + 7 × 0.1 + 8 × 0.6 = 0.6 + 0.7 + 4.8 = 6.1.
💡 How to solve: Expected Value & Binomial Probability
Expected value: E(X) = Σ x·P(x), multiply each outcome by its probability and add. Binomial: with n independent trials and success probability p, P(X = k) = C(n, k)·pk·(1 - p)n-k, and the mean is np.
Try a similar one
The table shows the probability distribution of X. Find the expected value E(X).
| x | 2 | 4 | 9 |
|---|---|---|---|
| P(x) | 0.4 | 0.2 | 0.4 |
✓ You've got it! Press Next to keep going. Not yet — the answer is . Read the steps above once more.
Not quite. The answer is
💡 You win $22 with probability 16 and lose $2 (an outcome of -2) with probability 56. E = 22 × 16 + (-2) × 56 = 22 − 106 = 126 = 2. The expected value is $2.
✓ Correct!
Not quite. The answer is
💡 You win $22 with probability 16 and lose $2 (an outcome of -2) with probability 56. E = 22 × 16 + (-2) × 56 = 22 − 106 = 126 = 2. The expected value is $2.
💡 How to solve: Expected Value & Binomial Probability
Expected value: E(X) = Σ x·P(x), multiply each outcome by its probability and add. Binomial: with n independent trials and success probability p, P(X = k) = C(n, k)·pk·(1 - p)n-k, and the mean is np.
Try a similar one
The table shows the probability distribution of X. Find the expected value E(X).
| x | 6 | 8 | 9 |
|---|---|---|---|
| P(x) | 0.1 | 0.4 | 0.5 |
✓ You've got it! Press Next to keep going. Not yet — the answer is . Read the steps above once more.
Not quite. The answer is
💡 Binomial probability with n = 3, k = 3 and p = 12: P(X = 3) = C(3, 3)·(12)3·(12)0 = C3, 323 = 18.
✓ Correct!
Not quite. The answer is
💡 Binomial probability with n = 3, k = 3 and p = 12: P(X = 3) = C(3, 3)·(12)3·(12)0 = C3, 323 = 18.
💡 How to solve: Expected Value & Binomial Probability
Expected value: E(X) = Σ x·P(x), multiply each outcome by its probability and add. Binomial: with n independent trials and success probability p, P(X = k) = C(n, k)·pk·(1 - p)n-k, and the mean is np.
Try a similar one
The table shows the probability distribution of X. Find the expected value E(X).
| x | 1 | 4 | 6 |
|---|---|---|---|
| P(x) | 0.3 | 0.1 | 0.6 |
✓ You've got it! Press Next to keep going. Not yet — the answer is . Read the steps above once more.
Not quite. The answer is
💡 For a binomial count the mean is np: 200 × 10% = 200 × 0.1 = 20. The expected number of made shots is 20.
✓ Correct!
Not quite. The answer is
💡 For a binomial count the mean is np: 200 × 10% = 200 × 0.1 = 20. The expected number of made shots is 20.
💡 How to solve: Expected Value & Binomial Probability
Expected value: E(X) = Σ x·P(x), multiply each outcome by its probability and add. Binomial: with n independent trials and success probability p, P(X = k) = C(n, k)·pk·(1 - p)n-k, and the mean is np.
Try a similar one
The table shows the probability distribution of X. Find the expected value E(X).
| x | 2 | 4 | 10 |
|---|---|---|---|
| P(x) | 0.3 | 0.4 | 0.3 |
✓ You've got it! Press Next to keep going. Not yet — the answer is . Read the steps above once more.
Not quite. The answer is
💡 For a binomial count the mean is np: 120 × 80% = 120 × 0.8 = 96. The expected number of made shots is 96.
✓ Correct!
Not quite. The answer is
💡 For a binomial count the mean is np: 120 × 80% = 120 × 0.8 = 96. The expected number of made shots is 96.
💡 How to solve: Expected Value & Binomial Probability
Expected value: E(X) = Σ x·P(x), multiply each outcome by its probability and add. Binomial: with n independent trials and success probability p, P(X = k) = C(n, k)·pk·(1 - p)n-k, and the mean is np.
Try a similar one
The table shows the probability distribution of X. Find the expected value E(X).
| x | 1 | 2 | 10 |
|---|---|---|---|
| P(x) | 0.2 | 0.4 | 0.4 |
✓ You've got it! Press Next to keep going. Not yet — the answer is . Read the steps above once more.
Not quite. The answer is
💡 For a binomial count the mean is np: 60 × 50% = 60 × 0.5 = 30. The expected number of made shots is 30.
✓ Correct!
Not quite. The answer is
💡 For a binomial count the mean is np: 60 × 50% = 60 × 0.5 = 30. The expected number of made shots is 30.
💡 How to solve: Expected Value & Binomial Probability
Expected value: E(X) = Σ x·P(x), multiply each outcome by its probability and add. Binomial: with n independent trials and success probability p, P(X = k) = C(n, k)·pk·(1 - p)n-k, and the mean is np.
Try a similar one
The table shows the probability distribution of X. Find the expected value E(X).
| x | 0 | 3 | 9 |
|---|---|---|---|
| P(x) | 0.1 | 0.3 | 0.6 |
✓ You've got it! Press Next to keep going. Not yet — the answer is . Read the steps above once more.
Not quite. The answer is
💡 For a binomial count the mean is np: 200 × 25% = 200 × 0.25 = 50. The expected number of made shots is 50.
✓ Correct!
Not quite. The answer is
💡 For a binomial count the mean is np: 200 × 25% = 200 × 0.25 = 50. The expected number of made shots is 50.
💡 How to solve: Expected Value & Binomial Probability
Expected value: E(X) = Σ x·P(x), multiply each outcome by its probability and add. Binomial: with n independent trials and success probability p, P(X = k) = C(n, k)·pk·(1 - p)n-k, and the mean is np.
Try a similar one
The table shows the probability distribution of X. Find the expected value E(X).
| x | 6 | 9 | 10 |
|---|---|---|---|
| P(x) | 0.3 | 0.1 | 0.6 |
✓ You've got it! Press Next to keep going. Not yet — the answer is . Read the steps above once more.
Not quite. The answer is
💡 Binomial probability with n = 3, k = 2 and p = 12: P(X = 2) = C(3, 2)·(12)2·(12)1 = C3, 223 = 38.
✓ Correct!
Not quite. The answer is
💡 Binomial probability with n = 3, k = 2 and p = 12: P(X = 2) = C(3, 2)·(12)2·(12)1 = C3, 223 = 38.
💡 How to solve: Expected Value & Binomial Probability
Expected value: E(X) = Σ x·P(x), multiply each outcome by its probability and add. Binomial: with n independent trials and success probability p, P(X = k) = C(n, k)·pk·(1 - p)n-k, and the mean is np.
Try a similar one
The table shows the probability distribution of X. Find the expected value E(X).
| x | 1 | 2 | 5 |
|---|---|---|---|
| P(x) | 0.3 | 0.4 | 0.3 |
✓ You've got it! Press Next to keep going. Not yet — the answer is . Read the steps above once more.
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Expected Value & Binomial Probability — Easy Worksheet 1
Solve each problem. Give probabilities as fractions in simplest form or as decimals rounded the way the problem says.