Understanding Independent Events - Statistics and Probability

  1. Identify whether AA and BB are independent if P(A)=25P(A)=\frac{2}{5}, P(B)=12P(B)=\frac{1}{2}, and P(A∩B)=15P(A\cap B)=\frac{1}{5}.
    Independent. Check: 25×12=15\frac{2}{5} \times \frac{1}{2} = \frac{1}{5}, which equals P(A∩B)P(A \cap B).
  2. Which statement is always true about independence: does AA being independent of BB imply BB independent of AA?
    Yes, independence is symmetric. If P(A∩B)=P(A)P(B)P(A\cap B)=P(A)P(B), then both directions hold.
  3. Identify whether AA and BB are independent if P(A)=13P(A)=\frac{1}{3}, P(B)=12P(B)=\frac{1}{2}, and P(A∩B)=16P(A\cap B)=\frac{1}{6}.
    Independent. 13×12=16\frac{1}{3} \times \frac{1}{2} = \frac{1}{6}, which equals P(A∩B)P(A\cap B).
  4. What equation tests independence using P(A∣B)P(A\mid B) when P(B)>0P(B)>0?
    P(A∣B)=P(A)P(A\mid B)=P(A). If BB doesn't affect AA's probability, they're independent.
  5. What is the independence condition using the product rule for events AA and BB?
    P(A∩B)=P(A)P(B)P(A \cap B)=P(A)P(B). Independent events satisfy this multiplication rule.
  6. Find P(B)P(B) if AA and BB are independent with P(A)=0.25P(A)=0.25 and P(A∩B)=0.05P(A\cap B)=0.05.
    0.20.2. Solve: 0.25×P(B)=0.050.25 \times P(B) = 0.05, so P(B)=0.2P(B) = 0.2.
  7. What is the independence test written using conditional probability P(B∣A)P(B\mid A)?
    P(B∣A)=P(B)P(B\mid A)=P(B). For independent events, knowing A occurred doesn't change the probability of B.
  8. Identify whether AA and BB are independent from this table: P(A)=0.4P(A)=0.4, P(B)=0.3P(B)=0.3, P(A∩B)=0.10P(A\cap B)=0.10.
    Not independent. 0.4×0.3=0.12≠0.100.4 \times 0.3 = 0.12 \neq 0.10.
  9. Find and correct the mistake: claiming independence because P(A∩B)=P(A)+P(B)P(A\cap B)=P(A)+P(B).
    Correct test: P(A∩B)=P(A)P(B)P(A\cap B)=P(A)P(B). Addition rule is for disjoint events, not independence.
  10. Find P(B)P(B) if AA and BB are independent, P(A)=0.25P(A)=0.25, and P(A∩B)=0.05P(A\cap B)=0.05.
    0.200.20. Solve 0.25×P(B)=0.050.25 \times P(B) = 0.05 to get P(B)=0.20P(B) = 0.20.
  11. Identify whether AA and BB are independent if P(A)=0.4P(A)=0.4, P(B)=0.5P(B)=0.5, and P(A∩B)=0.2P(A\cap B)=0.2.
    Independent. 0.4×0.5=0.20.4 \times 0.5 = 0.2, which equals P(A∩B)P(A\cap B).
  12. Identify whether AA and BB are independent if P(A)=25P(A)=\frac{2}{5}, P(B)=12P(B)=\frac{1}{2}, and P(A∩B)=14P(A\cap B)=\frac{1}{4}.
    Not independent. Check: 25×12=15≠14\frac{2}{5} \times \frac{1}{2} = \frac{1}{5} \neq \frac{1}{4}.
  13. Identify whether AA and BB are independent if P(A)=0.3P(A)=0.3, P(B)=0.6P(B)=0.6, and P(A∩B)=0.25P(A\cap B)=0.25.
    Not independent. 0.3×0.6=0.18≠0.250.3 \times 0.6 = 0.18 \neq 0.25.
  14. What is the defining equation for independence using the intersection probability?
    P(A∩B)=P(A)P(B)P(A \cap B)=P(A)P(B). Events are independent when their joint probability equals the product of individual probabilities.
  15. Find P(A∩B)P(A\cap B) if AA and BB are independent with P(A)=34P(A)=\frac{3}{4} and P(B)=23P(B)=\frac{2}{3}.
    12\frac{1}{2}. Multiply: 34×23=612=12\frac{3}{4} \times \frac{2}{3} = \frac{6}{12} = \frac{1}{2}.
  16. Find P(A)P(A) if AA and BB are independent, P(B)=0.8P(B)=0.8, and P(A∩B)=0.12P(A\cap B)=0.12.
    0.150.15. Solve P(A)×0.8=0.12P(A) \times 0.8 = 0.12 to get P(A)=0.15P(A) = 0.15.
  17. Find P(B∣A)P(B\mid A) if AA and BB are independent and P(B)=0.12P(B)=0.12.
    0.120.12. For independent events, P(B∣A)=P(B)P(B \mid A) = P(B).
  18. What is the symmetric conditional form of independence between events AA and BB?
    P(A∣B)=P(A)P(A\mid B)=P(A) and P(B∣A)=P(B)P(B\mid A)=P(B). Independence means each event's probability is unchanged by the other.
  19. Identify whether AA and BB are independent if P(A)=0.6P(A)=0.6, P(B)=0.3P(B)=0.3, and P(A∩B)=0.25P(A\cap B)=0.25.
    Not independent. Check: 0.6×0.3=0.18≠0.250.6 \times 0.3 = 0.18 \neq 0.25.
  20. Find P(A∣B)P(A\mid B) if AA and BB are independent and P(A)=0.35P(A)=0.35.
    0.350.35. For independent events, P(A∣B)=P(A)P(A \mid B) = P(A).
  21. Identify whether AA and BB are independent if P(A)=0.5P(A)=0.5, P(B)=0.4P(B)=0.4, and P(A∣B)=0.5P(A\mid B)=0.5.
    Independent. Since P(A∣B)=P(A)=0.5P(A \mid B) = P(A) = 0.5, they are independent.
  22. Which formula gives conditional probability P(B∣A)P(B\mid A) in terms of P(A∩B)P(A\cap B) and P(A)P(A)?
    P(B∣A)=P(A∩B)P(A)P(B\mid A)=\frac{P(A\cap B)}{P(A)}. Conditional probability is the ratio of joint to marginal probability.
  23. Find P(A∣B)P(A\mid B) if AA and BB are independent and P(A)=0.65P(A)=0.65.
    0.650.65. For independent events, P(A∣B)=P(A)P(A\mid B) = P(A).
  24. Which statement correctly describes independence: P(A∩B)>P(A)P(B)P(A\cap B)>P(A)P(B), ==, or <<?
    P(A∩B)=P(A)P(B)P(A\cap B)=P(A)P(B). Independence requires equality, not greater than or less than.
  25. Identify whether AA and BB are independent if P(A)=0.5P(A)=0.5 and P(A∣B)=0.5P(A\mid B)=0.5 with P(B)>0P(B)>0.
    Independent. P(A∣B)=P(A)P(A\mid B) = P(A) confirms independence.
  26. Which statement is true if AA and BB are independent: P(A∩B)P(A\cap B) equals what expression?
    P(A∩B)=P(A)P(B)P(A\cap B)=P(A)P(B). This is the defining property of independent events.
  27. What is P(A∩B)P(A \cap B) if AA and BB are independent and you know P(A)P(A) and P(B)P(B)?
    P(A∩B)=P(A)P(B)P(A \cap B)=P(A)P(B). For independent events, multiply their individual probabilities.
  28. Find P(A)P(A) if AA and BB are independent with P(B)=0.8P(B)=0.8 and P(A∩B)=0.24P(A\cap B)=0.24.
    0.30.3. Solve: P(A)×0.8=0.24P(A) \times 0.8 = 0.24, so P(A)=0.3P(A) = 0.3.
  29. Which probability must equal P(A)P(B)P(A)P(B) for events AA and BB to be independent?
    P(A∩B)P(A \cap B). The intersection probability must equal the product for independence.
  30. What equation tests independence using P(B∣A)P(B\mid A) when P(A)>0P(A)>0?
    P(B∣A)=P(B)P(B\mid A)=P(B). If AA doesn't affect BB's probability, they're independent.
  31. Identify whether AA and BB are independent if P(A)=0.5P(A)=0.5, P(B)=0.4P(B)=0.4, and P(A∣B)=0.6P(A\mid B)=0.6.
    Not independent. Since P(A∣B)=0.6≠0.5=P(A)P(A \mid B) = 0.6 \neq 0.5 = P(A), they're not independent.
  32. What is the equivalent independence condition using P(A∩B)P(A\cap B) and P(B)P(B) when P(B)>0P(B)>0?
    P(A∣B)=P(A∩B)P(B)=P(A)P(A\mid B)=\frac{P(A\cap B)}{P(B)}=P(A). Rearranging the conditional probability formula shows independence.
  33. Find P(A∩B)P(A\cap B) if AA and BB are independent with P(A)=0.7P(A)=0.7 and P(B)=0.2P(B)=0.2.
    0.140.14. Multiply: 0.7×0.2=0.140.7 \times 0.2 = 0.14.
  34. What is the independence test written using conditional probability P(A∣B)P(A\mid B)?
    P(A∣B)=P(A)P(A\mid B)=P(A). For independent events, knowing B occurred doesn't change the probability of A.
  35. Find P(A∩B)P(A\cap B) if AA and BB are independent, P(A)=0.7P(A)=0.7, and P(B)=0.2P(B)=0.2.
    0.140.14. For independent events, multiply: 0.7×0.2=0.140.7 \times 0.2 = 0.14.
  36. Identify whether AA and BB are independent if P(A)=0.5P(A)=0.5 and P(A∣B)=0.7P(A\mid B)=0.7 with P(B)>0P(B)>0.
    Not independent. P(A∣B)≠P(A)P(A\mid B) \neq P(A) means not independent.
  37. Which formula gives conditional probability P(A∣B)P(A\mid B) in terms of P(A∩B)P(A\cap B) and P(B)P(B)?
    P(A∣B)=P(A∩B)P(B)P(A\mid B)=\frac{P(A\cap B)}{P(B)}. Conditional probability is the ratio of joint to marginal probability.

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