A navigation briefcase contains 15 identical-looking USB drives. 9 of the drives contain m… — Armed Services Vocational Aptitude Battery (ASVAB)

A navigation briefcase contains $15$ identical-looking USB drives. $9$ of the drives contain map data, and $6$ 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.

Cevap: 12/35 / 12 / 35

Cevap

12/35

The correct answer is derived by multiplying the initial probability of drawing a map drive (

\frac{9}{15}

) by the updated probability of drawing a second map drive from the remaining inventory (

\frac{8}{14}

). Simplifying the fractions to

\frac{3}{5}

and

\frac{4}{7}

before multiplying gives the final result of

\frac{12}{35}

Adım Adım Çözüm

Determine the probability of selecting a map data drive on the first draw.

P(First Map) =

9

map drives /

15

total drives =

\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.

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) =

8

map drives remaining /

14

total drives remaining =

\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.

Calculate the compound probability of both dependent events occurring.

\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.

Anahtar Kavram

Compound probability of dependent events (without replacement)

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