Mathematics, 25.10.2019 04:43 shayyy49

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.

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Answer from: jalenclarke25

There is a 33.10% probability that there is at least one defective integrated circuit.

Step-by-step explanation:

For either integrated circuit, there are only two possible outcomes. Either they are defective, or they are not. This means that we can solve this problem using the binomial probability distribution.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

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

In which $C_{n,x}$ is the number of different combinatios of x objects from a set of n elements, given by the following formula.

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

And $\pi$ is the probability of X happening.

In this problem

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

The probability that any integrated circuit is defective is 0.01, so $\pi = 0.01$

What is the probability that there is at least one defective integrated circuit?

Either there is at least one defective integrated circuit, that is probability P(X 0) , or there are no defective integrated circuits, that is probability P(X = 0) . The sum of these probabilities is decimal 1. We want to find P(X0) .

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

There is a 33.10% probability that there is at least one defective integrated circuit.

Answer from: ccarwile01

0.33

Step-by-step explanation:

There are 2 possible outcomes. It's either the integrated circuit is defective or not defective

The binomial distribution can be used

P(x=r) = nCr× q^n-r×p^r 1

Where n =Number of integrated circuits = 40

p=The probability that any integrated circuit is defective is 0.01

q=The probability that any integrated circuit is not defective is 0.01

Probability of at least one defective =probability of defective being more than or equal to 1. This also equals 1 - probability of defective being less than 1 which equals 1 - probability of defective being 0

From eqn 1

P(x=0) = 40C0 × 0.99^40-0×0.01^0

= 1×0.669×1

=0.669

Probability of at least one defective = 1-0.669= 0.331 =

0.33 to 2 decimal places

Answer from: Quest

step-by-step explanation: