Question

Suppose that, for every lot of 100 computer chips a company produces, an average of 1.4 are defective. Another company buys many lots of these chips at a time, from which one lot is selected randomly and tested for defects. If the tested lot contains more than three defects, the buyer will reject all the lots sent in that batch. What is the probability that the buyer will accept the lots

Answer 1

Answer:

$P(X\leq 3)= P(X=0) +P(X=1)+P(X=2) +P(X=3)$

And we can find the individual probabilities like this:

$P(X=0) =1.4^0 \frac{e^{-1.4}}{0!}= 0.2466$

$P(X=1) =1.4^1 \frac{e^{-1.4}}{1!}= 0.3452$

$P(X=2) =1.4^2 \frac{e^{-1.4}}{2!}= 0.2417$

$P(X=3) =1.4^3 \frac{e^{-1.4}}{3!}= 0.1128$

And replacing we got:

$P(X\leq 3)= P(X=0) +P(X=1)+P(X=2) +P(X=3)=0.2466+ 0.3452+0.2417+0.1128=0.9463$

Step-by-step explanation:

Definitions and concepts

The Poisson process is useful when we want to analyze the probability of ocurrence of an event in a time specified. The probability distribution for a random variable X following the Poisson distribution is given by:

$P(X=x) =\lambda^x \frac{e^{-\lambda}}{x!}$

And the parameter $\lambda=1.4$ represent the average ocurrence rate per unit of time.

Solution to the problem

For this case the batch would be rejected if we found more than 3 defects, so then the probability of accept the batch would be given by:

$P(X\leq 3)= P(X=0) +P(X=1)+P(X=2) +P(X=3)$

And we can find the individual probabilities like this:

$P(X=0) =1.4^0 \frac{e^{-1.4}}{0!}= 0.2466$

$P(X=1) =1.4^1 \frac{e^{-1.4}}{1!}= 0.3452$

$P(X=2) =1.4^2 \frac{e^{-1.4}}{2!}= 0.2417$

$P(X=3) =1.4^3 \frac{e^{-1.4}}{3!}= 0.1128$

And replacing we got:

$P(X\leq 3)= P(X=0) +P(X=1)+P(X=2) +P(X=3)=0.2466+ 0.3452+0.2417+0.1128=0.9463$