Skip to main content

Problem 100: A shipment of 120 burglar alarms contains 5 that are defective. If 3 of these alarms are randomly selected and shipped to a customer, find the probability that the customer will get one bad unit by using
a. the formula for the hypergeometric distribution;
b. the formula for the binomial distribution as an approximation.

Problem 100: A shipment of 120 burglar alarms contains 5 that are defective. If 3 of these alarms are randomly selected and shipped to a customer, find the probability that the customer will get one bad unit by using
a. the formula for the hypergeometric distribution;
b. the formula for the binomial distribution as an approximation.


Solution:


a. the formula for the hypergeometric distribution
Given that $x=1,n=3,a=5, N=120$
Hypergeometric distribution, $h(x;n,a,N)=\frac{\left(\begin{array}{c}a\\ x\end{array}\right)\left(\begin{array}{c}N-a\\ n-x\end{array}\right)}{\left(\begin{array}{c}N\\ n\end{array}\right)}$
$P(x=1) = h(1;3,5,120)=\frac{\left(\begin{array}{c}5\\ 1\end{array}\right)\left(\begin{array}{c}120-5\\ 3-1\end{array}\right)}{\left(\begin{array}{c}120\\ 3\end{array}\right)} = 0.1167.$

b. the formula for the binomial distribution as an approximation
Given that $x=1,n=3,p = \frac{5}{120}$
$P(x= 1) = b(1;3,\frac{5}{120})$
$b(x;n,p)=\left(\begin{array}{c}n\\ x\end{array}\right){p}^{x}(1-p)^{n-x}$
$P(x= 1) = b(1;3,\frac{5}{120}) = \left(\begin{array}{c}3\\ 1\end{array}\right){\frac{5}{120}}^{1}(1-\frac{5}{120})^{3-1} = 0.1148$

Comments

Popular posts from this blog

Problem 98: If 6 of 18 new buildings in a city violate the building code, what is the probability that a building inspector, who randomly selects 4 of the new buildings for inspection, will catch
a. none of the buildings that violate the building code?
b. 1 of the new buildings that violate the building code?
c. 2 of the new buildings that violate the building code?
d. at least 3 of the new buildings that violate the building code?

Problem 98: If 6 of 18 new buildings in a city violate the building code, what is the probability that a building inspector, who randomly selects 4 of the new buildings for inspection, will catch a. none of the buildings that violate the building code? b. 1 of the new buildings that violate the building code? c. 2 of the new buildings that violate the building code? d. at least 3 of the new buildings that violate the building code? Solution: a. none of the buildings that violate the building code? Given that $x=0,n=4,a=6, N=18$ Hypergeometric distribution, $h(x;n,a,N)=\frac{\left(\begin{array}{c}a\\ x\end{array}\right)\left(\begin{array}{c}N-a\\ n-x\end{array}\right)}{\left(\begin{array}{c}N\\ n\end{array}\right)}$ $P(x=0) = h(0;4,6,18)=\frac{\left(\begin{array}{c}6\\ 0\end{array}\right)\left(\begin{array}{c}18-6\\ 4-0\end{array}\right)}{\left(\begin{array}{c}18\\ 4\end{array}\right)} = 0.16176.$ b. 1 of the new buildings that violate the building code? Given ...

Problem 105: If a bank receives on the average α = 6 bad checks per day, what are the probabilities that it will receive (a) 4 bad checks on any given day?
(b) 10 bad checks over any 2 consecutive days?

Problem 105: If a bank receives on the average α = 6 bad checks per day, what are the probabilities that it will receive (a) 4 bad checks on any given day? (b) 10 bad checks over any 2 consecutive days? Solution: Poisson distribution, $f(x;\lambda )=\frac{{\lambda }^{x}{e}^{-\lambda }}{x!}$ a. 4 bad checks on any given day Given that $x=4, \lambda = \alphaT = 6\dot 1 = 6$ $f(4;6 )=\frac{{6 }^{4}{e}^{-6 }}{4!} = 0.134$ b. 10 bad checks over any 2 consecutive days Given that $x=10, \lambda = \alphaT = 6\dot 2 = 12$ $f(10;12 )=\frac{{12 }^{10}{e}^{-12 }}{10!} = 0.104837$

Problem 57: Determine whether it is a subspace of the given vector space $\V$
$\V = \R^2$, and $S$ is the set of all vectors $(x, y)$ in $\V$ satisfying $3x + 2y = 0$.

Problem 57: Determine whether it is a subspace of the given vector space $\V$ $\V = \R^2$, and $S$ is the set of all vectors $(x, y)$ in $\V$ satisfying $3x + 2y = 0$. Solution: $S = \{x \in \R^2 : 3x+2y=0\}$. Let $(x_1,x_2),(y_1,y_2) \in S$. Then $3x_1+2x_2 = 0$ and $3y_1+2y_2 = 0$ Hence, $x + y = (x_1,x_2)+(y_1,y_2) =3x_1+2x_2 + 3y_1+2y_2 = 3(x_1+y_1) + 2(x_2+y_2)$ which implies that $(x_1+y_1,x_2+y_2) \in S$ Consequently, $S$ is closed under addition. Let $a \in \R$ and $(x_1,x_2) \in S$, then $ax = a(x_1,x_2) = a(3x_1+2x_2) = a0 = 0$ $=3(ax_1) + 2(ax_2) =0 $, which implies that $(ax_1,ax_2) \in S$ Therefore $S$ is also closed under scalar multiplication. It follows that $S$ is a subspace of $\R^2$.