English Deutsch Français Italiano Español Português 繁體中文 Bahasa Indonesia Tiếng Việt ภาษาไทย
All categories

A manufacturer makes two models of an item: model I, which accounts for 80% of unit sales, and model II, which accounts for 20% of unit sales. Because of defects, the manufacturer has to replace 10% of its model I and 18% of its model II. If a model is selected at random, find the probability that it will be defective.

I need to know how to get to the answer most importantly.

2006-10-15 14:29:07 · 7 answers · asked by Anonymous in Science & Mathematics Mathematics

7 answers

P(Model 1) = 0.8
P(Model 2) = 0.2
P(Defect|Model 1) = 0.1
P(Defect|Model 2) = 0.18

Below "or" can be interpretted as a union and "and" as an intersection.

P(Defect) = P{(Defect and Model 1) or (Defect and Model 2)}
= P(Defect and Model 1) + P(Defect and Model 2) (These events are mutually exclusive)
= P(Model 1)P(Defect|Model 1) + P(Model 2)P(Defect|Model 2) (This is the General Multiplication Rule.)
= 0.8*0.1 + 0.2*0.18

2006-10-15 14:37:05 · answer #1 · answered by blahb31 6 · 0 0

Model 1 = 80% of total items
Model 2 = 20% of total items

You need to figure out what 10% of 80% is 80/100 * 10/100 = 8%
Then 18% of 20% is 20/100 * 18/100 = 3.6%
Now you have that 8% of the total are defective model 1 items and 3.6% of the total are defective model 2 items. Now add the %-ages. 8% + 3.6% is 11.6% chance that one will be defective if selected at random.

2006-10-15 14:35:33 · answer #2 · answered by Mary Catherine M 2 · 0 0

Imagine there are 100 units.

80 of those units will be model 1.

20 will be model 2.

Of those 80, 10% will be defective, so 8 out of 80 are bad.

Of the 20, 18% will be bad, or 3.6 units...we'll call it 4 units are bad. So, 4/20 are bad.

So, out of 100 units...8 model 1, and 4 model 2 are bad.

8 + 4 = 12

12/100 = 12% chance of selecting a bad unit.

Regards,

Mysstere

2006-10-15 14:33:06 · answer #3 · answered by mysstere 5 · 1 1

Probability is 11.6%.
Assume that manufacturer makes 1000 of the famous widgets. 800 are type I, 200 are type II. Of the 800 TI, 80 are defective (on the average); of the 200 TII, 36 are defective. Thus, there are 116 defective widgets, or 11.6% of the total.

2006-10-15 14:39:46 · answer #4 · answered by Anonymous · 0 0

i choose the ten undesirable--SO i assumed this one outONEOUT AND MY answer IS.........27% probability OF GETTIN THE ALL HEARTS IN sequence FROM AN ACE to 10. that's HOW I ARRIVED at my decision, ok???? hear properly reason i basically madeit to point 3 after too long of a time and now i'm attempting point 4. if i had fifty two enjoying cards that are in a deck of enjoying cards and u dealt me 5 like u suggested u did--that could go away me with forty seven enjoying cards. minus a 10 might equivalent 37percentand the percentages on that are even worse-r than that so upload yet another or sub-song yet another 10 for the percentages bein against me and that i basically ended up with a 27% and am i ceremony or rong??? i'm a kia

2016-12-26 20:14:02 · answer #5 · answered by Anonymous · 0 0

10%of 80=8, 18%of 20=3.6. Add these together and you get 11.6 out of 100, which is a probability of 0.116.

2006-10-15 14:33:25 · answer #6 · answered by zee_prime 6 · 0 0

then the sale formula will be model one = model one *80%, model two = model two * 20%, total sale = model one + model two
( model one * %80 + model two * %20 ) = total sale

because of defect we need to subtract some: the lost of model one will be model one lost = model one * 10%, the lost of model two lost = model two * 18%

formula is total sale = ( model one - model one lost ) + ( model two - model two lost )

final formula will be total sale = ( model one * 80% - model one * 80% * 10%) + ( model two * 20% - model two * 20% * 18% )

2006-10-15 14:48:03 · answer #7 · answered by Jackie...boy 1 · 0 0

fedest.com, questions and answers