Calculation of Probabilities from Odds
1. The probability of event A, P(A), is determined from the odds in favor of A, which are 2:1. This means that for every 2 favorable outcomes, there is 1 unfavorable outcome. Therefore, P(A)=2+12=32.
2. The probability of the union of events A and B, P(A∪B), is determined from the odds in favor of A∪B, which are 3:1. This means that for every 3 favorable outcomes, there is 1 unfavorable outcome. Therefore, P(A∪B)=3+13=43. [1]
Derivation of Bounds for P(B)
1. The formula for the probability of the union of two events is given by P(A∪B)=P(A)+P(B)−P(A∩B).
2. Substituting the known probabilities, the equation becomes 43=32+P(B)−P(A∩B).
3. Rearranging the equation to express P(A∩B) in terms of P(B) yields P(A∩B)=P(B)+32−43.
4. To simplify the expression, a common denominator for the fractions is found, which is 12. Thus, P(A∩B)=P(B)+128−129=P(B)−121.
5. The probability of the intersection of two events, P(A∩B), must satisfy the inequality 0≤P(A∩B)≤P(A).
6. Substituting the expression for P(A∩B) and the value of P(A) into the inequality results in 0≤P(B)−121≤32.
7. The first part of the inequality, 0≤P(B)−121, implies P(B)≥121.
8. The second part of the inequality, P(B)−121≤32, implies P(B)≤32+121.
9. To simplify the sum, a common denominator is used: P(B)≤128+121=129=43.
10. Combining both inequalities, the range for P(B) is established as 121≤P(B)≤43. [1, 2]
Final Answer
The smallest and largest values for the probability of event B are 121 and 43, respectively. Therefore, the correct option is (a) 121≤P(B)≤43.
