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Difficulty: HardComplementary Probability and At-Least-One Scenarios

A financial firm's investment committee consists of 8 senior analysts and 4 junior analysts. If a project review panel of 3 members is chosen at random without replacement from this committee, what is the probability that the panel contains at least one junior analyst?

  1. A
    1455\frac{14}{55}
  2. B
    2855\frac{28}{55}
  3. C
    1927\frac{19}{27}
  4. 4155\frac{41}{55}Answer
  5. E
    4755\frac{47}{55}

Answer

The probability that the panel contains at least one junior analyst is 4155\frac{41}{55}.
To find the probability of selecting at least one junior analyst, it is most efficient to use the complementary probability rule: P(at least one junior)=1−P(no junior)P(\text{at least one junior}) = 1 - P(\text{no junior}). The total number of ways to choose 3 panel members from 12 committee members is (123)=220\binom{12}{3} = 220. The number of ways to choose 3 senior analysts from 8 is (83)=56\binom{8}{3} = 56. Thus, the probability of selecting no junior analysts is 56220=1455\frac{56}{220} = \frac{14}{55}. Subtracting this from 1 gives 1−1455=41551 - \frac{14}{55} = \frac{41}{55}.

Step-by-Step Solution

1
Calculate total possible outcomes for choosing 3 members out of 12.
(123)=12×11×103×2×1=220\binom{12}{3} = \frac{12 \times 11 \times 10}{3 \times 2 \times 1} = 220 total combinations.
Selection order does not matter, so combinations nCrnCr are used.
2
Calculate the number of unfavorable outcomes where zero junior analysts are chosen (all 3 selected are senior analysts).
(83)=8×7×63×2×1=56\binom{8}{3} = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = 56 ways.
All 3 members must be selected exclusively from the 8 senior analysts.
3
Find the probability of selecting zero junior analysts.
P(no junior analysts)=56220=1455P(\text{no junior analysts}) = \frac{56}{220} = \frac{14}{55}.
Divide the unfavorable outcomes by the total outcomes.
4
Apply the complementary probability principle: P(at least 1 junior)=1−P(no junior)P(\text{at least 1 junior}) = 1 - P(\text{no junior}).
P(at least 1 junior)=1−1455=4155P(\text{at least 1 junior}) = 1 - \frac{14}{55} = \frac{41}{55}.
The event 'at least one junior analyst' is the logical complement of 'no junior analysts'.

Key Concept

Complementary Probability
Estimated Time:2m 0s
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Topics

Complementary Probability and At-Least-One Scenarios

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