A testing kit contains 8 electronic sensors, of which exactly 3 are uncalibrated and 5 are… — GMAT

A testing kit contains $8$ electronic sensors, of which exactly $3$ are uncalibrated and $5$ are fully calibrated. If $3$ sensors are chosen at random without replacement, what is the probability that at least $1$ of the chosen sensors is uncalibrated?

A
$\frac{5}{28}$
B
$\frac{15}{28}$
$\frac{23}{28}$ Answer
D
$\frac{55}{56}$
E
$\frac{387}{512}$

Answer

The correct answer is

\frac{23}{28}

To find the probability that at least one sensor is uncalibrated, calculate 1 minus the probability that no selected sensors are uncalibrated. Out of 8 sensors, there are

\binom{8}{3} = 56

ways to pick 3. The number of ways to pick 3 calibrated sensors from the 5 available is

\binom{5}{3} = 10

. Thus,

P(\text{no uncalibrated}) = \frac{10}{56} = \frac{5}{28}

. Subtracting this from 1 gives

1 - \frac{5}{28} = \frac{23}{28}

Step-by-Step Solution

Calculate the total number of ways to choose 3 sensors out of 8.

\binom{8}{3} = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = 56

total outcomes.

Selection is made without replacement and order does not matter.

Calculate the complementary event: selecting 0 uncalibrated sensors (all 3 selected sensors are calibrated).

\binom{5}{3} = \binom{5}{2} = \frac{5 \times 4}{2 \times 1} = 10

outcomes with all calibrated sensors.

There are 5 calibrated sensors available, and 3 must be selected from them.

Find the probability of the complementary event (no uncalibrated sensors selected).

P(\text{no uncalibrated}) = \frac{10}{56} = \frac{5}{28}

Divide the favorable outcomes for selecting all calibrated sensors by total possible outcomes.

Apply the complementary probability principle to find

P(\text{at least one uncalibrated})

P(\text{at least one uncalibrated}) = 1 - P(\text{no uncalibrated}) = 1 - \frac{5}{28} = \frac{23}{28}

The sum of probabilities of complementary events is 1.

Key Concept

Complementary Probability: P(at least one X) = 1 - P(no X)

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