Chapter 4.4, Problem 15E

a.

To determine

Find the probability that the numbers of orders for small, medium and large drinks are 5, 7, and 8 respectively.

Answer to Problem 15E

The probability that the numbers of orders for small, medium and large drinks are 5, 7, and 8 respectively is 0.0411.

Explanation of Solution

Given info:

Total number of orders selected at random for audit is 20. There is about 25% of drink orders are for a small drink, 35% for a medium and 40% for a large.

Calculation:

The random variables $X_1,X_2,X_3$ are defined as the number of orders for small, medium and large drinks, respectively. Then $X_1,X_2,X_3$ follows Multinomial distribution with $n=20,p_1\text{=0}\text{.25},p_2\text{=0}\text{.35 and }{p}_3\text{=0}\text{.4 }$

Multinomial distribution:

If the random variables $X_1,\mathrm{...},X_k$ follows multinomial distribution with parameters $n,p_1,p_2,\mathrm{...},p_k$, then the probability mass function of the random variables is,

$\begin{array}{c}p\left(x_1,\mathrm{...},x_k\right)=P\left(X_1=x_1,\mathrm{...},X_k=x_k\right)\\ =\frac{n!}{x_1!x_2!\mathrm{...}{x}_k!}{p}_1{}^{x_1}{p}_2{}^{x_2}\mathrm{...}{p}_k{}^{x_k}{x}_i=0,1,2,\mathrm{...},n\text{ and }{\displaystyle \sum x_i}=n\end{array}$

Substitute $n=20,p_1\text{=0}\text{.25},p_2\text{=0}\text{.35 and }{p}_3\text{=0}\text{.4 }$ in the formula.

$\begin{array}{c}P\left(X_1=5,X_2=7\text{ and }{X}_3=8\right)=\frac{20!}{5!7!8!}{\left(0.25\right)}^5{\left(0.35\right)}^7{\left(0.4\right)}^8\\ =\left(\begin{array}{c}\frac{2.4329\times {10}^{18}}{\left(120\right)\left(5,040\right)\left(40,320\right)}\left(9.7656\times {10}^{-4}\right)\left(6.4339\times {10}^{-4}\right)\\ \left(6.5536\times {10}^{-4}\right)\end{array}\right)\\ =\left(99,768,240\right)\left(4.1177\times {10}^{-10}\right)\\ =0.0411\end{array}$

Thus, the probability that the numbers of orders for small, medium and large drinks are 5, 7, and 8 respectively is 0.0411.

b.

To determine

Find the probability that more than 10 orders are for large drinks.

Answer to Problem 15E

The probability that more than 10 orders are for large drinks is 0.1275.

Explanation of Solution

Calculation:

If the random variables $X_1,\mathrm{...},X_k$ follows multinomial distribution with parameters $n,p_1,p_2,\mathrm{...},p_k$, then $X_i$ follows binomial with parameters $n\text{ and }{p}_i$ for each i.

Binomial distribution:

The probability of obtaining x successes in n independent trails of a binomial experiment is,

$P\left(X=x\right)=\left(\begin{array}{l}n\\ x\end{array}\right)p^x{\left(1-p\right)}^{n-x},x=0,1,2,\mathrm{...},n$

Where, p is the probability of success.

$P\left({\text{X}}_3>10\right)=\left(\begin{array}{l}P\left(X_3=11\right)+P\left(X_3=12\right)+P\left(X_3=13\right)+P\left(X_3=14\right)+P\left(X_3=15\right)\\ P\left(X_3=16\right)+P\left(X_3=17\right)+P\left(X_3=18\right)+P\left(X_3=19\right)+P\left(X_3=20\right)\end{array}\right)$

Substitute x as 11,12,13,14,15,16,17,18,19,20 and p as 0.4in the formula.

$\begin{array}{c}P\left({\text{X}}_3>10\right)=\left(\begin{array}{c}\frac{20!}{11!\left(20-11\right)!}{\left(0.4\right)}^{11}{\left(1-0.4\right)}^{20-11}+\frac{20!}{12!\left(20-12\right)!}{\left(0.4\right)}^{12}{\left(1-0.4\right)}^{20-12}\\ +\frac{20!}{13!\left(20-13\right)!}{\left(0.4\right)}^{13}{\left(1-0.4\right)}^{20-13}+\frac{20!}{14!\left(20-14\right)!}{\left(0.4\right)}^{14}{\left(1-0.4\right)}^{20-14}\\ +\frac{20!}{15!\left(20-15\right)!}{\left(0.4\right)}^{15}{\left(1-0.4\right)}^{20-15}+\frac{20!}{16!\left(20-16\right)!}{\left(0.4\right)}^{16}{\left(1-0.4\right)}^{20-16}\\ +\frac{20!}{17!\left(20-17\right)!}{\left(0.4\right)}^{17}{\left(1-0.4\right)}^{20-17}+\frac{20!}{18!\left(20-18\right)!}{\left(0.4\right)}^{18}{\left(1-0.4\right)}^{20-18}\\ +\frac{20!}{19!\left(20-19\right)!}{\left(0.4\right)}^{19}{\left(1-0.4\right)}^{20-19}+\frac{20!}{20!\left(20-20\right)!}{\left(0.4\right)}^{20}{\left(1-0.4\right)}^{20-20}\end{array}\right)\\ =\left(\begin{array}{c}\frac{2.4329\times {10}^{18}}{\left(39,916,800\right)\left(362,880\right)}\left(4.194\times {10}^{-5}\right)\left(0.0101\right)\\ +\frac{2.4329\times {10}^{18}}{\left(479,001,600\right)\left(40,320\right)}\left(1.6777\times {10}^{-5}\right)\left(0.0167\right)\\ +\frac{2.4329\times {10}^{18}}{\left(6,227,020,800\right)\left(5,040\right)}\left(6.7108\times {10}^{-6}\right)\left(0.0279\right)\\ +\frac{2.4329\times {10}^{18}}{\left(8.7178\times {10}^{10}\right)\left(720\right)}\left(2.6843\times {10}^{-6}\right)\left(0.0466\right)\\ +\frac{2.4329\times {10}^{18}}{\left(1.3077\times {10}^{12}\right)\left(120\right)}\left(1.0737\times {10}^{-6}\right)\left(0.0777\right)\\ +\frac{2.4329\times {10}^{18}}{\left(2.0923\times {10}^{13}\right)\left(24\right)}\left(4.2949\times {10}^{-7}\right)\left(0.1296\right)\\ +\frac{2.4329\times {10}^{18}}{\left(3.5569\times {10}^{14}\right)\left(6\right)}\left(1.7179\times {10}^{-7}\right)\left(0.216\right)\\ ++\frac{2.4329\times {10}^{18}}{\left(6.4024\times {10}^{15}\right)\left(2\right)}\left(6.8719\times {10}^{-8}\right)\left(0.36\right)\\ +\frac{2.4329\times {10}^{18}}{\left(1.2164\times {10}^{17}\right)\left(1\right)}\left(2.7487\times {10}^{-8}\right)\left(0.6\right)\\ +\frac{2.4329\times {10}^{18}}{\left(2.4329\times {10}^{18}\right)\left(1\right)}\left(1.0995\times {10}^{-8}\right)\left(1\right)\end{array}\right)\\ =\left(\begin{array}{c}\left(167,960\right)\left(4.2269\times {10}^{-7}\right)+\left(125,970\right)\left(2.8179\times {10}^{-7}\right)\\ +\left(77,520\right)\left(1.8786\times {10}^{-7}\right)+\left(38,760\right)\left(1.2524\times {10}^{-7}\right)\\ +\left(15,504\right)\left(8.3494\times {10}^{-8}\right)+\left(4,845\right)\left(5.5663\times {10}^{-8}\right)\\ +\left(1,140\right)\left(3.7108\times {10}^{-8}\right)+\left(190\right)\left(2.4739\times {10}^{-8}\right)\\ +\left(20\right)\left(1.6493\times {10}^{-8}\right)+\left(1\right)\left(1.0995\times {10}^{-8}\right)\end{array}\right)\\ =\left(\begin{array}{c}0.0709+0.0355+0.0146+0.0048+0.0013+0.0003+0.00004\\ +0.000004+0.0000003+0.00000001\end{array}\right)\end{array}$

                  $=0.1275$

Thus, the probability that more than 10 orders are for large drinks is 0.1275.