1. The probability of getting a type A blood donor is 0.4, and the probability of getting a type B blood donor is 0.2. What is the probability that the first two donors are both type A?
- Solution: Since the events are independent, multiply the probabilities: 0.4 × 0.4 = 0.16.
2. What is the probability that the first donor is type A and the second is type B?
- Solution: Multiply the individual probabilities: 0.4 × 0.2 = 0.08.
3. What is the probability that the first donor is type B and the second is type A?
- Solution: Multiply the individual probabilities: 0.2 × 0.4 = 0.08.
4. What is the probability that the first two donors are different blood types?
- Solution: Add the probabilities of the mutually exclusive cases (A then B, or B then A): 0.08 + 0.08 = 0.16.
5. What is the probability that at least one of the first two donors is type A?
- Solution: Use the complement rule. Probability neither is type A = 0.6 × 0.6 = 0.36. So, probability at least one is type A = 1 - 0.36 = 0.64.
6. If the third donor is also independent and has a 0.4 probability of being type A, what is the probability that all three are type A?
- Solution: Multiply the probabilities: 0.4 × 0.4 × 0.4 = 0.064.
7. What is the probability that exactly two out of the first three donors are type A?
- Solution: There are three possible sequences: AAB, ABA, BAA. Each has probability 0.4 × 0.4 × 0.6 = 0.096. Total probability = 3 × 0.096 = 0.288.
8. What is the probability that at least two out of the first three donors are type A?
- Solution: Add the probabilities of exactly two type A and exactly three type A: 0.288 + 0.064 = 0.352.
Parent Tip: Review the logic above to help your child master the concept of codominance worksheet.
