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I went with the complement approach: probability of zero reds is (7/12)*(6/11) which gives 42/132, so at least one red is 1 minus that, about 0.682.
Use the complement rule: find the probability of drawing no red balls (i.e., two blue balls) and subtract from 1. Alternatively, compute the probability of at least one red directly by summing the cases of one red and two reds. Clearly state the sample space and that draws are without replacement.
Pro tip: Mention that the complement method is often simpler and less error-prone, especially when the event 'at least one' is involved. Also, note that the answer can be verified by computing the direct probability and comparing.
Identify that there are 5 red and 7 blue balls, total 12. Two balls are drawn without replacement. We need P(at least one red).
Decide between direct calculation (sum of P(1 red) and P(2 red)) or complement rule (1 - P(0 red)). The complement is usually easier.
Calculate P(no red) = P(both blue) = (7/12) * (6/11) = 42/132 = 7/22.
Subtract from 1: P(at least one red) = 1 - 7/22 = 15/22.
Optionally verify by direct calculation: P(1 red) = (5/12)*(7/11) + (7/12)*(5/11) = 70/132; P(2 red) = (5/12)*(4/11) = 20/132; sum = 90/132 = 15/22. Present answer clearly.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
First, clarify the problem setup: ask whether the balls are drawn with or without replacement and the initial counts of blue and red balls. Then, apply conditional probability using the formula P(second red | first blue) = P(first blue and second red) / P(first blue), computing each probability based on the sampling method.
Pro tip: Explicitly state your assumptions about the urn composition and sampling method before calculating—this shows attention to detail and prevents misinterpreting the problem. If the counts aren't given, use variables and express the answer in terms of them.
Ask for the initial number of blue and red balls, and whether the first ball is replaced before the second draw. Confirm if the draw is random and each ball is equally likely.
Let A be the event that the first ball is blue, and B be the event that the second ball is red. We need P(B|A).
Calculate P(A) and P(A ∩ B) based on the sampling method. For without replacement, P(A) = B/(B+R) and P(A ∩ B) = (B/(B+R)) * (R/(B+R-1)). For with replacement, P(A) = B/(B+R) and P(A ∩ B) = (B/(B+R)) * (R/(B+R)).
Use P(B|A) = P(A ∩ B) / P(A). Simplify the expression to get the final probability.
Present the final probability, clearly stating any assumptions made (e.g., without replacement). If variables were used, provide the formula.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
Used linearity of expectation here, which is the right move.
First, clarify the urn composition and whether draws are with or without replacement. Then use linearity of expectation to compute the expected number of red balls as the sum of the expected reds on each draw, handling dependence if sampling without replacement.
Pro tip: Mention that linearity of expectation holds even when draws are dependent, which is a key insight that simplifies the problem. Also, briefly discuss how the answer changes if the urn composition is unknown, showing awareness of probabilistic nuances.
Ask for the urn's composition (number of red and non-red balls) and whether draws are with or without replacement. State any assumptions if not provided.
Let X_i be 1 if the i-th draw is red, 0 otherwise. The total number of red balls is X_1 + X_2.
Compute E[X_1] and E[X_2] separately. For the first draw, E[X_1] = R/(R+B). For the second draw, if without replacement, E[X_2] = R/(R+B) as well (by symmetry or direct calculation).
The expected total is E[X_1] + E[X_2] = 2R/(R+B) for both with and without replacement (when the urn has R red and B blue balls).
Mention that if the urn composition is unknown, the expected value depends on the prior distribution. Also note that the result generalizes to n draws: n * R/(R+B).
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.