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Difficulty: HardIndependent and Dependent Events

A research laboratory's secure vault contains NN data drives, where N>4N > 4. Exactly 33 of the drives contain encrypted security keys, while the remaining N−3N - 3 drives contain standard operational data. If two data drives are selected at random one after another without replacement, the probability that at least one of the selected drives contains an encrypted security key is 815\frac{8}{15}. What is the value of NN?

  1. A
    8
  2. B
    9
  3. 10Answer
  4. D
    12
  5. E
    15

Answer

10
The correct option is 10. The probability of choosing at least one encrypted security key drive in two draws without replacement is 1−P(no security key drives)1 - P(\text{no security key drives}). Since there are N−3N - 3 standard drives out of NN total drives, P(no security key drives)=N−3N×N−4N−1P(\text{no security key drives}) = \frac{N - 3}{N} \times \frac{N - 4}{N - 1}. Setting 1−(N−3)(N−4)N(N−1)=8151 - \frac{(N - 3)(N - 4)}{N(N - 1)} = \frac{8}{15} gives (N−3)(N−4)N(N−1)=715\frac{(N - 3)(N - 4)}{N(N - 1)} = \frac{7}{15}. Solving 15(N2−7N+12)=7(N2−N)15(N^2 - 7N + 12) = 7(N^2 - N) leads to 8N2−98N+180=08N^2 - 98N + 180 = 0, or 4N2−49N+90=04N^2 - 49N + 90 = 0, which factors as (4N−9)(N−10)=0(4N - 9)(N - 10) = 0. Because NN must be an integer, N=10N = 10.

Step-by-Step Solution

1
Express the complementary probability event
The probability of at least one drive having an encrypted key is 1−P(both drives are standard data drives)1 - P(\text{both drives are standard data drives}). Thus, P(both standard)=1−815=715P(\text{both standard}) = 1 - \frac{8}{15} = \frac{7}{15}.
Using the complementary probability rule is more efficient than calculating the individual probabilities of key-then-standard, standard-then-key, and key-then-key.
2
Set up the probability expression for dependent sequential selection without replacement
For the first draw, the probability of selecting a standard drive is N−3N\frac{N - 3}{N}. For the second draw, since one standard drive has been removed, the probability is N−4N−1\frac{N - 4}{N - 1}. Therefore, (N−3)(N−4)N(N−1)=715\frac{(N - 3)(N - 4)}{N(N - 1)} = \frac{7}{15}.
The events are dependent because the selection is made without replacement.
3
Solve the algebraic equation for N
Expanding both sides gives 15(N2−7N+12)=7(N2−N)15(N^2 - 7N + 12) = 7(N^2 - N), which simplifies to 15N2−105N+180=7N2−7N  ⟹  8N2−98N+180=015N^2 - 105N + 180 = 7N^2 - 7N \implies 8N^2 - 98N + 180 = 0. Dividing by 2 yields 4N2−49N+90=04N^2 - 49N + 90 = 0. Factoring gives (4N−9)(N−10)=0(4N - 9)(N - 10) = 0.
Since NN must be a positive integer greater than 4, N=10N = 10.

Key Concept

Dependent probability without replacement and complementary probability
Estimated Time:2m 0s
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Topics

Independent and Dependent Events

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