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Conditional Probability & Independent Events — Easy Worksheet 1
Grade 11 · MathSheets Name: ____________________ Date: __________
Write each probability as a fraction in simplest form, like 3/8. P(A | B) means the probability of A given that B has happened.
How to solve
Independent events (one does not change the other): P(A and B) = P(A) × P(B). Dependent events, such as drawing without replacement: multiply by the probability of the second event after the first has happened. Addition rule: P(A or B) = P(A) + P(B) - P(A and B); for mutually exclusive events P(A and B) = 0, so just add. Conditional probability: P(A | B) = P(A and B) ÷ P(B). In a two-way table, look only at the row or column for B and divide the A-and-B count by that total. A and B are independent exactly when P(A and B) = P(A) × P(B).
Example
A bag has 2 red and 3 blue marbles. One marble is drawn and put back, then a second marble is drawn. Find the probability that the first is blue and the second is red.
Answer: 625
Not quite. The answer is
💡 There are 9 marbles. P(first is red) = 39. The marble is put back, so the draws are independent and P(second is blue) = 69. Multiply: 39 × 69 = 1881 = 29.
✓ Correct!
Not quite. The answer is
💡 There are 9 marbles. P(first is red) = 39. The marble is put back, so the draws are independent and P(second is blue) = 69. Multiply: 39 × 69 = 1881 = 29.
💡 How to solve: Conditional Probability & Independent Events
Independent events (one does not change the other): P(A and B) = P(A) × P(B). Dependent events, such as drawing without replacement: multiply by the probability of the second event after the first has happened. Addition rule: P(A or B) = P(A) + P(B) - P(A and B); for mutually exclusive events P(A and B) = 0, so just add. Conditional probability: P(A | B) = P(A and B) ÷ P(B). In a two-way table, look only at the row or column for B and divide the A-and-B count by that total. A and B are independent exactly when P(A and B) = P(A) × P(B).
Try a similar one
A bag has 6 red and 6 blue marbles. One marble is drawn and put back, then a second marble is drawn. Find the probability that the first is blue and the second is red.
✓ 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
💡 Mutually exclusive events cannot happen together, so P(A and B) = 0 and P(A or B) = P(A) + P(B) = 14 + 58 = 28 + 58 = 78.
✓ Correct!
Not quite. The answer is
💡 Mutually exclusive events cannot happen together, so P(A and B) = 0 and P(A or B) = P(A) + P(B) = 14 + 58 = 28 + 58 = 78.
💡 How to solve: Conditional Probability & Independent Events
Independent events (one does not change the other): P(A and B) = P(A) × P(B). Dependent events, such as drawing without replacement: multiply by the probability of the second event after the first has happened. Addition rule: P(A or B) = P(A) + P(B) - P(A and B); for mutually exclusive events P(A and B) = 0, so just add. Conditional probability: P(A | B) = P(A and B) ÷ P(B). In a two-way table, look only at the row or column for B and divide the A-and-B count by that total. A and B are independent exactly when P(A and B) = P(A) × P(B).
Try a similar one
A bag has 2 red and 4 blue marbles. One marble is drawn and put back, then a second marble is drawn. Find the probability that both marbles are blue.
✓ You've got it! Press Next to keep going. Not yet — the answer is . Read the steps above once more.
| Job | No job | Total | |
|---|---|---|---|
| Juniors | 10 | 10 | 20 |
| Seniors | 8 | 12 | 20 |
| Total | 18 | 22 | 40 |
Not quite. The answer is
💡 Look only at the Seniors row: 8 + 12 = 20 in all. Of these, 8 are in the Job column. Using counts, P(A | B) = P(A and B) ÷ P(B) = (840) ÷ (2040) = 8 ÷ 20 = 820 = 25.
✓ Correct!
Not quite. The answer is
💡 Look only at the Seniors row: 8 + 12 = 20 in all. Of these, 8 are in the Job column. Using counts, P(A | B) = P(A and B) ÷ P(B) = (840) ÷ (2040) = 8 ÷ 20 = 820 = 25.
💡 How to solve: Conditional Probability & Independent Events
Independent events (one does not change the other): P(A and B) = P(A) × P(B). Dependent events, such as drawing without replacement: multiply by the probability of the second event after the first has happened. Addition rule: P(A or B) = P(A) + P(B) - P(A and B); for mutually exclusive events P(A and B) = 0, so just add. Conditional probability: P(A | B) = P(A and B) ÷ P(B). In a two-way table, look only at the row or column for B and divide the A-and-B count by that total. A and B are independent exactly when P(A and B) = P(A) × P(B).
Try a similar one
A bag has 3 red and 6 blue marbles. One marble is drawn and put back, then a second marble is drawn. Find the probability that both marbles are blue.
✓ 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
💡 There are 11 marbles. P(first is blue) = 511. The marble is put back, so the draws are independent and P(second is blue) = 511. Multiply: 511 × 511 = 25121.
✓ Correct!
Not quite. The answer is
💡 There are 11 marbles. P(first is blue) = 511. The marble is put back, so the draws are independent and P(second is blue) = 511. Multiply: 511 × 511 = 25121.
💡 How to solve: Conditional Probability & Independent Events
Independent events (one does not change the other): P(A and B) = P(A) × P(B). Dependent events, such as drawing without replacement: multiply by the probability of the second event after the first has happened. Addition rule: P(A or B) = P(A) + P(B) - P(A and B); for mutually exclusive events P(A and B) = 0, so just add. Conditional probability: P(A | B) = P(A and B) ÷ P(B). In a two-way table, look only at the row or column for B and divide the A-and-B count by that total. A and B are independent exactly when P(A and B) = P(A) × P(B).
Try a similar one
A bag has 5 red and 6 blue marbles. One marble is drawn and put back, then a second marble is drawn. Find the probability that both marbles are red.
✓ 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
💡 There are 9 marbles. P(first is blue) = 39. The marble is put back, so the draws are independent and P(second is red) = 69. Multiply: 39 × 69 = 1881 = 29.
✓ Correct!
Not quite. The answer is
💡 There are 9 marbles. P(first is blue) = 39. The marble is put back, so the draws are independent and P(second is red) = 69. Multiply: 39 × 69 = 1881 = 29.
💡 How to solve: Conditional Probability & Independent Events
Independent events (one does not change the other): P(A and B) = P(A) × P(B). Dependent events, such as drawing without replacement: multiply by the probability of the second event after the first has happened. Addition rule: P(A or B) = P(A) + P(B) - P(A and B); for mutually exclusive events P(A and B) = 0, so just add. Conditional probability: P(A | B) = P(A and B) ÷ P(B). In a two-way table, look only at the row or column for B and divide the A-and-B count by that total. A and B are independent exactly when P(A and B) = P(A) × P(B).
Try a similar one
A bag has 5 red and 2 blue marbles. One marble is drawn and put back, then a second marble is drawn. Find the probability that both marbles are blue.
✓ 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
💡 Mutually exclusive events cannot happen together, so P(A and B) = 0 and P(A or B) = P(A) + P(B) = 18 + 14 = 18 + 28 = 38.
✓ Correct!
Not quite. The answer is
💡 Mutually exclusive events cannot happen together, so P(A and B) = 0 and P(A or B) = P(A) + P(B) = 18 + 14 = 18 + 28 = 38.
💡 How to solve: Conditional Probability & Independent Events
Independent events (one does not change the other): P(A and B) = P(A) × P(B). Dependent events, such as drawing without replacement: multiply by the probability of the second event after the first has happened. Addition rule: P(A or B) = P(A) + P(B) - P(A and B); for mutually exclusive events P(A and B) = 0, so just add. Conditional probability: P(A | B) = P(A and B) ÷ P(B). In a two-way table, look only at the row or column for B and divide the A-and-B count by that total. A and B are independent exactly when P(A and B) = P(A) × P(B).
Try a similar one
A bag has 4 red and 6 blue marbles. One marble is drawn and put back, then a second marble is drawn. Find the probability that the first is blue and the second is red.
✓ You've got it! Press Next to keep going. Not yet — the answer is . Read the steps above once more.
| Liked it | Did not like it | Total | |
|---|---|---|---|
| Adults | 8 | 2 | 10 |
| Teens | 6 | 11 | 17 |
| Total | 14 | 13 | 27 |
Not quite. The answer is
💡 Look only at the Teens row: 6 + 11 = 17 in all. Of these, 6 are in the Liked it column. Using counts, P(A | B) = P(A and B) ÷ P(B) = (627) ÷ (1727) = 6 ÷ 17 = 617.
✓ Correct!
Not quite. The answer is
💡 Look only at the Teens row: 6 + 11 = 17 in all. Of these, 6 are in the Liked it column. Using counts, P(A | B) = P(A and B) ÷ P(B) = (627) ÷ (1727) = 6 ÷ 17 = 617.
💡 How to solve: Conditional Probability & Independent Events
Independent events (one does not change the other): P(A and B) = P(A) × P(B). Dependent events, such as drawing without replacement: multiply by the probability of the second event after the first has happened. Addition rule: P(A or B) = P(A) + P(B) - P(A and B); for mutually exclusive events P(A and B) = 0, so just add. Conditional probability: P(A | B) = P(A and B) ÷ P(B). In a two-way table, look only at the row or column for B and divide the A-and-B count by that total. A and B are independent exactly when P(A and B) = P(A) × P(B).
Try a similar one
A bag has 2 red and 6 blue marbles. One marble is drawn and put back, then a second marble is drawn. Find the probability that both marbles are blue.
✓ 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
💡 There are 10 marbles. P(first is red) = 410. The marble is put back, so the draws are independent and P(second is blue) = 610. Multiply: 410 × 610 = 24100 = 625.
✓ Correct!
Not quite. The answer is
💡 There are 10 marbles. P(first is red) = 410. The marble is put back, so the draws are independent and P(second is blue) = 610. Multiply: 410 × 610 = 24100 = 625.
💡 How to solve: Conditional Probability & Independent Events
Independent events (one does not change the other): P(A and B) = P(A) × P(B). Dependent events, such as drawing without replacement: multiply by the probability of the second event after the first has happened. Addition rule: P(A or B) = P(A) + P(B) - P(A and B); for mutually exclusive events P(A and B) = 0, so just add. Conditional probability: P(A | B) = P(A and B) ÷ P(B). In a two-way table, look only at the row or column for B and divide the A-and-B count by that total. A and B are independent exactly when P(A and B) = P(A) × P(B).
Try a similar one
A bag has 5 red and 4 blue marbles. One marble is drawn and put back, then a second marble is drawn. Find the probability that both marbles are blue.
✓ You've got it! Press Next to keep going. Not yet — the answer is . Read the steps above once more.
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Conditional Probability & Independent Events — Easy Worksheet 1
Write each probability as a fraction in simplest form, like 3/8. P(A | B) means the probability of A given that B has happened.