Probability that bulb is defective questions

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SUMMARY

This discussion focuses on calculating probabilities related to defective light bulbs and parts from vendors. The first problem involves determining the probability of selecting exactly 2 out of 10 bulbs being 40-watt from a box of 100 bulbs, where 20 are 40-watt. The second problem addresses the defect rates of parts supplied by two vendors, A and B, with respective defect rates of 0.01 and 0.03. Key calculations include using combinations (nCr) to find probabilities and understanding the relationship between defect rates and total parts supplied.

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Homework Statement


There are 100 light bulbs in a box. 20 are 40 watt, the rest are 60 watt.
If we randomly choose 10 bulbs, what is the probability that there will be exactly 2 40-watt bulbs?

What is the probability that in a random sample of 10 bulbs there will be at least 1 40-watt bulb?

and then i am also having trouble with this one


75% of a certain part is supplied by vendor A and 25% by vendor B. Vendor A's defect rate is 0.01 and vendor B's defect rate is 0.03.

If a part is from vendor A, what is the probability it is defective?

What is the probability that a randomly chosen part will be defective and from vendor A?

What is the probability that a randomly chosen part will be non-defective and from Vendor B?

The Attempt at a Solution



I think that these are nCr problems but I am very confused by these problems, if someone could explain them to me that would be amazing, thank you so much.
 
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schapman22 said:

Homework Statement


There are 100 light bulbs in a box. 20 are 40 watt, the rest are 60 watt.
If we randomly choose 10 bulbs, what is the probability that there will be exactly 2 40-watt bulbs?
The probability that the first is 40 watt is, of course, 20/100= .2. That would leave 99 bulbs with 19 of them 40 watt. The probability that the second bulb is also 40 watt is 19/99 so that both first and second will be 40 watt is (20/100)(19/99). That leaves 98 bulbs of which 80 are NOT 40 watt so the probability the third is NOT 40 watt is 80/98, the probabiility the fourth is not 40 watt is 79/97, the probability the fifth is not 40 watt is 78/96, etc. That is the probability of the first two being 40 Watt and the other 8 not is (20/100)(19/99)(80/98)(79/97)...(73/91).

If you apply the same analysis to, say, "4th and 6th bulbs 40 watt all others not" or "3rd and 9th bulbs 40 watt others not", etc. you should soon realize that you are just moving numbers in the numerator around and, in fact, get the same number repeatedly so you really just have to multiply that first number by the number of ways to put 2 40 watt bulbs and 8 non- 40 watt bulbs which is the "binomial coefficient" 2C10.

What is the probability that in a random sample of 10 bulbs there will be at least 1 40-watt bulb?
This is actually easier. "At least one 40-wtt bulb" is the opposite of "all 60 watt" and so is 1 minus the probability that all 10 are 60 watt.

and then i am also having trouble with this one75% of a certain part is supplied by vendor A and 25% by vendor B. Vendor A's defect rate is 0.01 and vendor B's defect rate is 0.03.

If a part is from vendor A, what is the probability it is defective?
This doesn't require any calculation- it is just given: "Vendor A's defect rte is 0.01"

What is the probability that a randomly chosen part will be defective and from vendor A?
I would imagine 10000 parts. We are told that 75% or 7500 are from vendor A and 1% of those, or 75 are defective. So the probability is 75/1000= .075.

What is the probability that a randomly chosen part will be non-defective and from Vendor B?
Same idea. If there were 1000 parts, 25% or 250 are from vendor B. 3% of those, 7.5 are defective so the probability is 7.5/1000.

The Attempt at a Solution



I think that these are nCr problems but I am very confused by these problems, if someone could explain them to me that would be amazing, thank you so much.
 
Last edited by a moderator:
Thank you so much, that was very helpful and cleared up some things.
 
My only question is for the first one you said to multiply .2* 2C10, however when i try to do 2C10 i get 0, or an error
 
schapman22 said:
My only question is for the first one you said to multiply .2* 2C10, however when i try to do 2C10 i get 0, or an error
Halls meant 10C2, the number of ways of choosing 2 items from 10.
... and he didn't say the answer was .2 * 10C2.
In terms of combinations, you have to choose 2 from the 10 and 8 from the 80. How many ways does that make?
That is compared with all the possible choices, namely, 10 from 100.
 
Last edited:
for 10C2 I'm geting 45, if i multiply 45*.2 I get 9, doesn't my answer have to be between 0 and 1?
 
My edit wasn't quite fast enough. See what I added in the previous post.
 

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