Question

consider the following binomial experiment: the probability that a cell phone manufactured at electronics unlimited is defective is 0.11. if a sample of 6 cell phones is selected at random, what is the probability that at least one is defective? a) 0.4423 b) 0.5030 c) 0.4970 d) 0.5584 e) 0.4416 f) none of the above.

Answer 1

Using the binomial distribution, it is found that there is a 0.5954 = 59.54% probability that at least 1 is defective.

What is the binomial distribution?

A collection of numerous independently distributed Bernoulli trials with equal distributions results in the binomial distribution. The experiment in a Bernoulli trial is said to be random and can only result in one of two outcomes: success or failure.

For each phone, there are only two possible outcomes, either it is defective, or it is not. The probability of a phone being defective is independent of any other phone, which means that the binomial distribution is used to solve this question.

Binomial probability distribution

$P(X = x) = C_n, x. p^{x.(1-p)^n-x}$

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

The parameters are:

x is the number of successes.

n is the number of trials.

p is the probability of success on a single trial.

In this problem:

0.11 probability of a single phone is defective, hence

6 phones are selected at random, hence

The probability that at least 1 is defective is:

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

In which

$P(X = x) = C_n, x. p^{x.(1-p)^n-x}$.

P(x = 0) = C₆,₀.(0.11)⁰ (0.86)⁶ = 0.4046

Then:

P(X >= 1) = 1 - P(X = 0) = 1 - 0.4046 = 0.5954

Hence, the 0.5954 = 59.54% probability that at least 1 is defective.

To learn more about the binomial distribution visit,

https://brainly.com/question/9325204

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