An electronic product contains 40 integrated circuits. The probability that any integrated circuit is defective is 0.01, and the

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3 months ago

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An electronic product contains 40 integrated circuits. The probability that any integrated circuit is defective is 0.01, and the

integrated circuits are independent. The product operates only if there are no defective integrated circuits. What is the probability that there is at least one defective integrated circuit? Round to two decimal places.

Mathematics

2 answers:

PIT_PIT [12.4K] · Answer 1 · 2026-02-19T05:31:25+03:00

Answer: 0.33

Stepwise explanation:

There are two possible outcomes for each integrated circuit: it is either defective or not.

The binomial distribution applies here.

P(x=r) = nCr × q^(n−r) × p^r --------------1

Where n = total number of integrated circuits = 40

p = probability that any integrated circuit is defective = 0.01

q = probability that any integrated circuit is not defective = 0.99

The probability of having at least one defective integrated circuit equals 1 minus the probability of having zero defective ones.

From equation 1,

P(x=0) = 40C0 × 0.99^40 × 0.01^0

= 1 × 0.669 × 1

= 0.669

Therefore, probability of one or more defects = 1 − 0.669 = 0.331, which rounds to 0.33 to two decimal places.

Zina [12.3K] · Answer 2 · 2026-02-19T05:33:49+03:00

Answer:

The chance that there is one or more defective integrated circuits is 33.10%.

Detailed solution:

Each integrated circuit can be either defective or not, which provides only two possible outcomes. As a result, this scenario is well-suited to be analyzed using the binomial probability distribution.

About the binomial distribution

This distribution calculates the likelihood of exactly x successes in n repeated trials where each trial has two possible results.

$P(X = x) = C_{n,x}.\pi^{x}.(1-\pi)^{n-x}$

Here, $C_{n,x}$ represents the count of combinations of x items selected from n elements, described by the formula:

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And $\pi$ is the probability of the event X occurring.

Applying it to the problem

The product contains 40 integrated circuits, so $n = 40$ .

The probability that an individual integrated circuit is defective is 0.01, so $\pi = 0.01$ .

Finding the probability of at least one defective circuit

There are two cases: either at least one integrated circuit is defective (probability $P(X > 0)$ ) or none are defective (probability $P(X = 0)$ ). Since probabilities sum to 1, we want to determine $P(X>0)$ .

$P(X > 0) + P(X = 0) = 1$

$P(X > 0) = 1 - P(X = 0)$

$P(X = x) = C_{n,x}.\pi^{x}.(1-\pi)^{n-x}$

$P(X = 0) = C_{40,0}.(0.01)^{0}.(0.99)^{40} = 0.6690$

$P(X > 0) = 1 - P(X = 0) = 1 - 0.6690 = 0.3310$

Hence, the probability of having one or more defective integrated circuits is 33.10%.