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Difficulty: MediumCompound Probability

A navigation briefcase contains 1515 identical-looking USB drives. 99 of the drives contain map data, and 66 contain communication protocols. If a pilot randomly selects two drives from the briefcase one after the other without replacing the first, what is the probability that both selected drives contain map data? Express your answer as a simplified fraction.

Answer: 12/35 / 12 / 35

Answer

12/35
The correct answer is derived by multiplying the initial probability of drawing a map drive (915\frac{9}{15}) by the updated probability of drawing a second map drive from the remaining inventory (814\frac{8}{14}). Simplifying the fractions to 35\frac{3}{5} and 47\frac{4}{7} before multiplying gives the final result of 1235\frac{12}{35}.

Step-by-Step Solution

1
Determine the probability of selecting a map data drive on the first draw.
P(First Map) = 99 map drives / 1515 total drives = 915=35\frac{9}{15} = \frac{3}{5}.
The probability of a single event is the number of favorable outcomes divided by the total number of possible outcomes.
2
Determine the probability of selecting a map data drive on the second draw, given that one map drive has already been removed.
P(Second Map) = 88 map drives remaining / 1414 total drives remaining = 814=47\frac{8}{14} = \frac{4}{7}.
Since the first drive is not replaced, both the number of map drives and the total number of drives decrease by one. This is a dependent event.
3
Calculate the compound probability of both dependent events occurring.
35×47=1235\frac{3}{5} \times \frac{4}{7} = \frac{12}{35}.
To find the probability of two dependent events happening in sequence, you multiply the probability of the first event by the conditional probability of the second event.

Key Concept

Compound probability of dependent events (without replacement)
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Topics

Compound Probability

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