Probability of Compound Events Quiz

Grade 7 · Math · Quiz

Calculate probabilities for compound events using multiplication rules, sample spaces, and tree diagrams.

Questions

1. A jar contains 5 green and 3 yellow candies. You draw one, eat it, then draw another. What type of events are these?

   1. Independent
   2. Complementary
   3. Dependent
   4. Mutually exclusive
   Show answer

   Answer: Dependent

   Eating the first candy changes the total number and composition in the jar, so the second draw is affected. These are dependent events.

2. You flip a coin and roll a number cube (1–6). What is P(tails and 3)?

   1. 1/12
   2. 1/6
   3. 1/3
   4. 1/4
   Show answer

   Answer: 1/12

   P(tails) = 1/2 and P(3) = 1/6. Independent, so P(tails and 3) = 1/2 × 1/6 = 1/12.

3. A bag has 4 red, 3 blue, and 3 green marbles. You pick one, replace it, pick again. What is P(red, then red)?

   1. 8/10
   2. 4/25
   3. 16/100
   4. 2/5
   Show answer

   Answer: 16/100

   P(red) = 4/10 = 2/5. With replacement, P(red then red) = 2/5 × 2/5 = 4/25 = 16/100. Both representations are correct; 16/100 is option C.

4. How many outcomes are in the sample space when you roll two dice?

   1. 12
   2. 18
   3. 36
   4. 24
   Show answer

   Answer: 36

   Each die has 6 faces. Total outcomes = 6 × 6 = 36.

5. You spin a spinner with sections 1, 2, 3, 4 (equal) and flip a coin. What is P(even number and heads)?

   1. 1/4
   2. 1/2
   3. 1/8
   4. 3/8
   Show answer

   Answer: 1/4

   Even numbers on the spinner are 2 and 4, so P(even) = 2/4 = 1/2. P(heads) = 1/2. P(even and heads) = 1/2 × 1/2 = 1/4.

6. A bag has 2 red and 8 blue marbles. Without replacement, what is P(red then red)?

   1. 1/25
   2. 2/90
   3. 4/90
   4. 1/45
   Show answer

   Answer: 2/90

   P(first red) = 2/10 = 1/5. After removing one red, P(second red) = 1/9. P(red then red) = 1/5 × 1/9 = 1/45 = 2/90.

7. The sum of all probabilities in a complete sample space equals 1.

   Show answer

   Answer: True

   All probabilities in a complete sample space must account for every possible outcome. Their sum is always 1 (or 100%).

8. P(A) = 1/3 and P(B) = 1/4. If A and B are independent, P(A and B) = _____.

   Show answer

   Answer: 1/12

   P(A and B) = P(A) × P(B) for independent events. 1/3 × 1/4 = 1/12.

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