Question

Problem 3 [Sans R y garbage or fake rubbish. We c xpectG0%of We run a website focusing on sclling either phon the users to buy the phony garbage versus the fake rubbish. Each customer is independent of cach other. Let be the number among the next 10 purchasers who buy phony garbage. (a). Compute the probability that r is exactly 10. (b). Compute the mean of r and the variance of z. (c). We only have 8 picces of phony garbage and 6 picces of fake rubbish. What is the probability that among the requests of these 10 customers can all be met from existing stock?

Answer 1

The probability that a randomly selected purchaser buys phony garbage is 0.60.

Let X be the number of customers out of 10 purchasers, who buy phony garbage. X has a Binomial distribution with parameters, number of trial, n=10 and the success probability p=0.60

The probability that X=x customers, out of 10 purchasers, buy phony garbage is given by

$\begin{align*} P(X=x)&=\binom{n}{x}p^x(1-p)^{n-x}\\ &=\binom{10}{x}0.60^x(1-0.60)^{10-x}\\ &=\frac{10!}{x!(10-x)!}0.60^x(1-0.60)^{10-x}\\ \end{align*}$

a) The probability that X=10 is

$\begin{align*} P(X=10)&=\frac{10!}{10!(10-10)!}0.60^{10}(1-0.60)^{10-10}\\ &=0.60^{10}\\ &=0.0060 \end{align*}$

b) The mean of X is (using the standard results of Binomial distribution)

$\begin{align*} E(X)=np=10\times 0.60=6 \end{align*}$

The Variance of X is (using the standard results of Binomial distribution)

$\begin{align*} Var(X)=np(1-p)=10\times 0.60\times (1-0.60)=2.4 \end{align*}$

c)We have 8 pieces of phony garbage and 6 pieces of fake garbage.

Out of the 10 purchases, we can sell at the most 8 pieces of phony garbage and we have to sell at least 10-6) = 4 pieces of phony garbage (so that the reset of 6 pieces can be fake garbage). the following table shows the amount that can be met before running out of stock

Phony garbage (X)	Fake Rubbish
4	6
5	5
6	4
7	3
8	2

That means the requests can be met as long as the value of X is between 4 and 8 (inclusive)

The probability that among the requests of these 10 customers can all be met from existing stock is

$\begin{align*} P(4\le X\le 8)&=P(X=4)+P(X=5)+...+P(X=8)\\ &=\frac{10!}{4!(10-4)!}0.60^{4}(1-0.60)^{10-4}+\frac{10!}{5!(10-5)!}0.60^{5}(1-0.60)^{10-5}+...+\frac{10!}{8!(10-8)!}0.60^{8}(1-0.60)^{10-8}\\ &=0.1115+ 0.2007+ 0.2508+ 0.215+ 0.1209\\ &=0.8989 \end{align*}$

ans: The probability that among the requests of these 10 customers can all be met from existing stock is 0.8989