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Difficulty: MediumIndependent and Dependent Events

A corporate compliance archive contains 1010 confidential files and 55 public files. An auditor randomly selects two files from the archive, one after another without replacement. What is the probability that at least one of the selected files is public?

  1. A
    221\frac{2}{21}
  2. B
    1021\frac{10}{21}
  3. C
    1121\frac{11}{21}
  4. D
    59\frac{5}{9}
  5. 47\frac{4}{7}Answer

Answer

The probability that at least one of the selected files is public is 47\frac{4}{7}.
To find the probability that at least one file is public, subtract the probability that no public files are selected (meaning both files are confidential) from 1. The probability of choosing a confidential file on the first draw is 1015=23\frac{10}{15} = \frac{2}{3}. Since the selection is without replacement, 99 confidential files remain out of 1414 total files. The probability of choosing a confidential file on the second draw is 914\frac{9}{14}. The joint probability of drawing two confidential files is 23×914=37\frac{2}{3} \times \frac{9}{14} = \frac{3}{7}. Subtracting this from 1 yields 1−37=471 - \frac{3}{7} = \frac{4}{7}.

Step-by-Step Solution

1
Identify the complementary event
The complementary event to 'at least one selected file is public' is 'both selected files are confidential'.
Calculating the probability of the complement is simpler than summing the probabilities of all positive individual outcomes.
2
Calculate the probability of drawing two confidential files sequentially without replacement
The probability of drawing a confidential file on the first selection is 1015=23\frac{10}{15} = \frac{2}{3}. After removing one confidential file, 9 confidential files remain out of 14 total files. The probability on the second selection is 914\frac{9}{14}. Thus, P(both confidential)=1015×914=23×914=37P(\text{both confidential}) = \frac{10}{15} \times \frac{9}{14} = \frac{2}{3} \times \frac{9}{14} = \frac{3}{7}.
Because the drawing is done without replacement, the outcome of the first draw affects the total sample space and event space for the second draw.
3
Subtract the complementary probability from 1
P(at least one public)=1−P(both confidential)=1−37=47P(\text{at least one public}) = 1 - P(\text{both confidential}) = 1 - \frac{3}{7} = \frac{4}{7}.
The sum of the probabilities of an event and its complement equals 1.

Key Concept

Probability of Dependent Events and Complementary Probability
Estimated Time:1m 30s
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Topics

Independent and Dependent Events

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