Mathematics, 29.07.2019 20:20 tiaragirl923

Abox contains 13 transistors, 4 of which are defective. if 4 are selected at random, find the probability that a. all are defective. b. none are defective.

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Answer from: bella2017

(a) The probability that all are defective is $\frac{1}{725}$ .

(b) The probability that all are defective is $\frac{126}{725}$ .

Step-by-step explanation:

It is given that a box contains 13 transistors, 4 of which are defective.

Total number of transistors = 13

Total number of defective transistors = 4

Total number of non-defective transistors = 13 - 4 = 9

Number of selected transistors = 4

Total possible ways to select 4 transistors from 13 transistors is

$\text{Possible outcomes}=^{13}C_4=\frac{13!}{4!(13-4)!}=725$

(a)

We need to find the probability that none are defective.

It means all 4 transistors are from defective transistors.

$\text{Required outcomes}=^4C_4=1$

The probability that all are defective is

$P=\frac{\text{Required outcomes}}{\text{Possible outcomes}}$

$P=\frac{1}{725}$

Therefore the probability that all are defective is $\frac{1}{725}$ .

(b)

We need to find the probability that none are defective.

It means all 4 transistors are from non-defective transistors.

$\text{Required outcomes}=^9C_4=\frac{9!}{4!(9-4)!}=126$

The probability that none are defective is

$P=\frac{\text{Required outcomes}}{\text{Possible outcomes}}$

$P=\frac{126}{725}$

Therefore the probability that none are defective is $\frac{126}{725}$ .

Answer from: Quest

see below

step-by-step explanation:

7. there are 10 data points in the set. the middle point is between 5 and 6. the middle point in the top half is 8 and the middle point in the bottom half is 3. to find the iqr , take point 8 and subtract point 3

mid term point 8 = 100 point 3 = 88

iqr = 100-88 = 12

final point 8 = 93 point 3 = 78

iqr = 93 - 78 = 15

the final has the greatest iqr

6. b - there is a high data value that causes the data set to be asymmetrical for the males: the data for males are high and asymmetrical . there is an outlier for the males that skews the data

5. the iqr is a better measure of spread for the movies and basketball games. since the basketball game has an outlier, we do not want to use standard deviation for measurement.

3. the exam median is much higher than the class median. the median line is farther to the right on the exams

2. there is no way to tell what the means are from the box and whisker plots.

Answer from: Quest

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step-by-step explanation:

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