Question

Three components are randomly sampled, one at a time, from a large lot. As each component is selected, it is tested. If it passes the test, a success (S) occurs; if it fails the test, a failure (F) occurs. Assume that 80% of the components in the lot will succeed in passing the test. Let X represent the number of successes among the three sampled components.
Find p(X=1). Find P(X=0). Find the mean of the random variable X and the interpret. Find the standard deviation of the random variable X. Let Y represent the number of successes if four components are sampled. Find P(Y=3). Let Z represent the number of successes if four components are sampled. Find P(Z=3).

Answer 1

Answer:

a) $P(X=0)=(3C0)(0.8)^0 (1-0.8)^{3-0}=0.008$

$P(X=1)=(3C1)(0.8)^1 (1-0.8)^{3-1}=0.096$  

$E(X) = np = 3*0.8= 2.4$

We expect 2.4 successes in a sample of three selected.

And the standard deviation is given by:

$Sd(X)= \sigma = \sqrt{np(1-p)}=\sqrt{3*0.8*(1-0.8)}= 0.693$

Represent the typical variation around the mean.

b) $Y \sim Binom(n=4, p=0.8)$

And we want this probability:

$P(Y=3)=(4C3)(0.8)^3 (1-0.8)^{4-3}=0.4096$

c) $Z \sim Binom(n=4, p=0.8)$

And we want this probability:

$P(Z=3)=(4C3)(0.8)^3 (1-0.8)^{4-3}=0.4096$

Step-by-step explanation:

Previous concepts

The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".

Solution to the problem

Part a

Let X the random variable of interest, on this case we now that:

$X \sim Binom(n=3, p=0.8)$

The probability mass function for the Binomial distribution is given as:

$P(X)=(nCx)(p)^x (1-p)^{n-x}$

Where (nCx) means combinatory and it's given by this formula:

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

And we want the following probabilities

$P(X=0)=(3C0)(0.8)^0 (1-0.8)^{3-0}=0.008$

$P(X=1)=(3C1)(0.8)^1 (1-0.8)^{3-1}=0.096$

The mean is given by:

$E(X) = np = 3*0.8= 2.4$

We expect 2.4 successes in a sample of three selected.

And the standard deviation is given by:

$Sd(X)= \sigma = \sqrt{np(1-p)}=\sqrt{3*0.8*(1-0.8)}= 0.693$

Represent the typical variation around the mean.

Part b

Let Y the random variable of interest, on this case we now that:

$Y \sim Binom(n=4, p=0.8)$

And we want this probability:

$P(Y=3)=(4C3)(0.8)^3 (1-0.8)^{4-3}=0.4096$

Part c

Let Z the random variable of interest, on this case we now that:

$Z \sim Binom(n=4, p=0.8)$

And we want this probability:

$P(Z=3)=(4C3)(0.8)^3 (1-0.8)^{4-3}=0.4096$