Question

A certain type of aluminum screen has, an average, one flaw in a 100-foot roll. Assume the flaw distribution approximately follows the Poisson distribution. (a) Find the probability that a 100-foor roll has at most 2 flaws. (b) Suppose that I bought 10 200-foot rolls, find the probability that there are exactly 3 rolls that have no flaws in them. (c) Suppose that a store had many 200-foot rolls in storage, a clerk is doing inspections. What is the probability that the 15th roll he inspected was the 3rd one that have no flaws in it

Answer 1

Answer:

a) 0.9197 = 91.97% probability that a 100-foor roll has at most 2 flaws.

b) 0.1074 = 10.74% probability that there are exactly 3 rolls that have no flaws in them.

c) 0.0394 = 3.94% probability that the 15th roll he inspected was the 3rd one that have no flaws in it.

Step-by-step explanation:

To solve this question, we need to understand the Poisson and the Binomial distribution.

Poisson distribution:

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

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

In which

x is the number of sucesses

e = 2.71828 is the Euler number

$\mu$ is the mean in the given interval.

Binomial probability distribution

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

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

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

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

And p is the probability of X happening.

(a) Find the probability that a 100-foor roll has at most 2 flaws.

A certain type of aluminum screen has, an average, one flaw in a 100-foot roll, which means that $\mu = 1$. Only one roll means that the Poisson distribution is used.

This is:

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

In which

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

$P(X = 0) = \frac{e^{-1}*1^{0}}{(0)!} = 0.3679$

$P(X = 1) = \frac{e^{-1}*1^{1}}{(1)!} = 0.3679$

$P(X = 2) = \frac{e^{-1}*1^{2}}{(2)!} = 0.1839$

$P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2) = 0.3679 + 0.3679 + 0.1839 = 0.9197$

0.9197 = 91.97% probability that a 100-foor roll has at most 2 flaws.

(b) Suppose that I bought 10 200-foot rolls, find the probability that there are exactly 3 rolls that have no flaws in them.

Two parts.

Probability that a single 200-foot roll has no flaw.

This is $P(X = 0)$, Poisson(single 200-foot roll) when $\mu = \frac{200*1}{100} = 2$. So

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

$P(X = 0) = \frac{e^{-2}*2^{0}}{(0)!} = 0.1353$

0.1353 probability that a single 200-foot roll has no flaw.

Probability that on 10 200-foot rolls, 3 have no flaws.

Multiple 200-foot rolls means that we use the binomial distribution.

0.1353 probability that a single 200-foot roll has no flaw means that $p = 0.1353$

10 200-foot rolls means that $n = 10$

We want $P(X = 3)$. So

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

$P(X = 3) = C_{10,3}.(0.1353)^{3}.(0.8647)^{7} = 0.1074$

0.1074 = 10.74% probability that there are exactly 3 rolls that have no flaws in them.

(c) What is the probability that the 15th roll he inspected was the 3rd one that have no flaws in it.

0.1353 probability that a single 200-foot roll has no flaw means that $p = 0.1353$

2 with flaws in the first 14, which is $P(X = 2)$ when $n = 14$

The 15th has no flaw, with probability of 0.1353. So

$P = 0.1353*P(X = 2) = 0.1353*(C_{14,2}.(0.1353)^{2}.(0.8647)^{12}) = 0.1353*0.2911 = 0.0394$

0.0394 = 3.94% probability that the 15th roll he inspected was the 3rd one that have no flaws in it