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A computer retail store has 15 personal computers in stock. A buyer wants to purchase 3 of them. Unknown to either the retail st

Dima020 [189]

Answer:

a. 1365 ways

b. Probability = 0.4096

c. Probability = 0.5904

Explanation:

Given

PCs = 15

Purchase = 3

Solving (a): Ways to select 4 computers out of 15, we make use of Combination formula as follows;

$^nC_r = \frac{n!}{(n-r)!r!}$

Where $n = 15\ and\ r = 4$

$^{15}C_4 = \frac{15!}{(15-4)!4!}$

$^{15}C_4 = \frac{15!}{11!4!}$

$^{15}C_4 = \frac{15 * 14 * 13 * 12 * 11!}{11! * 4 * 3 * 2 * 1}$

$^{15}C_4 = \frac{15 * 14 * 13 * 12}{4 * 3 * 2 * 1}$

$^{15}C_4 = \frac{32760}{24}$

$^{15}C_4 = 1365$

Hence, there are 1365 ways 

Solving (b): The probability that exactly 1 will be defective (from the selected 4)

First, we calculate the probability of a PC being defective (p) and probability of a PC not being defective (q)

From the given parameters; 3 out of 15 is detective;

So;

$p = 3/15$

$p = 0.2$

$q = 1 - p$

$q = 1 - 0.2$

$q = 0.8$

Solving further using binomial;

$(p + q)^n = p^n + ^nC_1p^{n-1}q + ^nC_2p^{n-2}q^2 + .....+q^n$

Where n = 4

For the probability that exactly 1 out of 4 will be defective, we make use of

$Probability = ^nC_3pq^3$

Substitute 4 for n, 0.2 for p and 0.8 for q

$Probability = ^4C_3 * 0.2 * 0.8^3$

$Probability = \frac{4!}{3!1!} * 0.2 * 0.8^3$

$Probability = 4 * 0.2 * 0.8^3$

$Probability = 0.4096$

Solving (c): Probability that at least one is defective;

In probability, opposite probability sums to 1;

Hence;

Probability that at least one is defective + Probability that at none is defective = 1

Probability that none is defective is calculated as thus;

$Probability = q^n$

Substitute 4 for n and 0.8 for q

$Probability = 0.8^4$

$Probability = 0.4096$

Substitute 0.4096 for Probability that at none is defective

Probability that at least one is defective + 0.4096= 1

Collect Like Terms

Probability = 1 - 0.4096

Probability = 0.5904

8 0

3 years ago

Drivers dealing with strong emotions may take risks they otherwise would not. true false

grigory [225]

This is true because some people may be going through rough times and when they aren't they probably won't feel the same way as they do when they go through tough times.

7 0

3 years ago

A data analyst is using the Color tool in Tableau to apply a color scheme to a data visualization. They want the visualization t

alina1380 [7]

Answer:

Color contrast is the difference in light between font (or anything in the foreground) and its background.

Explanation:

In web accessibility, how well one color stands out from another color determines whether or not most people will be able to read the information.

Contrast makes things look different and stand out

6 0

3 years ago

From the ages of 18 to 24, individuals born from 1980 to 1984 held an average of 6.2 jobs—which is slightly higher than the numb

Romashka [77]

Answer:

The parameters mentioned show how millennials have a much more marked tendency to job instability, that is, to change jobs more frequently than previous generations.

This is explained by two fundamental factors: on the one hand, the millennial generation differs from previous generations, especially because of their constant need for change and their greater emotional instability, which has a decisive influence on their constant and abrupt changes in their daily routines. , including their jobs; on the other, the deregulation of labor markets worldwide, which, on the one hand, have reduced the percentages of the unemployed population on the planet, but on the other hand, have given employers greater facilities to fire their employees, with which they they must change jobs more quickly.

3 0

3 years ago

Consider the following declaration: (1, 2) double currentBalance[91]; In this declaration, identify the following: a. The array

Rom4ik [11]

Answer:

Following are the solution to the given choices:

Explanation:

Given:

double currentBalance[91];//defining a double array

In point a:

The name of the array is= currentBalance.

In point b:

91 double values could be saved in the array. It requires 8bytes to hold a double that if the array size is 91*8 = 728

In point c:

Each element's data type is double.

In point d:

The array index range of values is between 0 and 90 (every array index starts from 0 and ends in N-1, here N=91).

In point e:

To access first element use currentBalance[0], for middle currentBalance[91/2] , for last currentBalance[90]

3 0

2 years ago