Binomial Distribution: fire resistance of office furniture

[Monkeyseat]

10) Safety inspectors carry out checks on the fire resistance of office furniture. Twenty percent of furniture checked failed the test.

a) Three randomly selected pieces of furniture are tested. Find the probability that:
i) none of these three items will fail the test

= 0.512
ii) exactly one of these three items will fail the test

= 0.384

b) Twenty pieces of office furniture awaiting test include exactly four which will fail. If three of these twenty pieces, selected at random are tested, find the probability that:
i) all three will past the test,
ii) exactly two will past the test.

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I am unsure how to do (b), please could someone help? Thanks. I've not been doing binomial distribution for long so go slowly, but can a cumulative binomial table be used for (b)? I had used them for all of the other questions in the exercise, but the binomial distribution tables I have only start at n=8. Please say if possible, thanks.

 

[soroban]

Re: Binomial Distribution

Hello, Monkeyseat!

For (b), there is a finite number of items.
. . We must use Combinations . . .

b) Twenty pieces of office furniture awaiting test include exactly four which will fail.

If three of these twenty pieces, selected at random are tested, find the probability that:

i) all three will past the test,
ii) exactly two will past the test.

\(\displaystyle \text{There are 20 objects: }\:\text{16 "Pass", 4 "Fail".}\)

\(\displaystyle \text{There are: }\;{20\choose3} \:=\:1140\text{ possible samples.}\)


\(\displaystyle \text{i) If all three Pass, there are: }\;{16\choose3} \:=\:560\text{ ways}\)

. . \(\displaystyle \text{Therefore: }\;P(\text{3 Pass}) \:=\:\frac{560}{1140} \:=\:\frac{28}{57}\)


\(\displaystyle \text{ii) If two Pass, then there is one Fail.}\)

\(\displaystyle \text{There are: }{16\choose2}{4\choose1} \:=\:480\text{ ways.}\)

. . \(\displaystyle \text{Therefore: }\;P(\text{2 Pass}) \:=\:\frac{480}{1140} \:=\:\frac{8}{19}\)


Edit: Correct error in part (ii)

 

[Monkeyseat]

Re: Binomial Distribution

Hi soroban, many thanks for your reply and thank you for taking the time to write such a detailed and straightforward response. It has been really helpful and I appreciate it a lot!

A few questions though, sorry:

1)

On ii) you say the answer is 480/560, would the answer of P(2 Pass & 1 Fail) not be 480/1140? This is what the answer in the book says. Can I also ask why (16C2) and (4C1) are multiplied? Is it (the number of ways of selecting 2 that work out of a total of 16 that work) * (the number of ways of selecting 1 that doesn't work out of a total of 4 that don't work)? Why multiply?

2)

I know your method is probably best/easiest but I was just wondering about what I found when I was trying to figure this out. Would the method shown below of doing the questions be acceptable:

i) (16/20) * (15/19) * (14/18) = 0.491 (3 s.f.)

ii) 3 * ((16/20) * (15/19) * (4/18)) = 0.421 (3 s.f.) <----- By the way, if it was done this way, do you multiply by 3 because of "(3C2)" or in other words being 3 ways to arrange them? Or is it (3C1)?

Can that method also be used for questions that can be done with binomial distribution? I think it worked when I altered it and tried it for (a). This is like the tree diagrams we had been doing in the past. I do see that the answers in (b) are approximate to the answers in (a) as they are proportional - still 20% probability in both. And by the way, because someone asked me last time, "(3 s.f.)" is just an abbreviation for 3 significant figures because I rounded it.

3)

So a binomial table/formula cannot be used because the events are dependent (i.e. the probability of the next outcome changes each time one is chosen?)? Is that what you mean by "finite"?

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Brackets everywhere hehe!

Many thanks. If you could just help with each of those points I'm unclear on it would be great. More assistance is appreciated. Sorry to set them out like this and ask so many questions, but it just makes it clearer for everyone on the bits I don't understand. I just like to make sure I understand things fully. Hope that's not too messy. Once again, cheers.

 

[soroban]

Re: Binomial Distribution

Hello, Monkeyseat

On ii) you say the answer is 480/560, would the answer of P(2 Pass & 1 Fail) not be 480/1140?

Yes, sorry . . . I've corrected my error in my post.

Why (16C2) and (4C1) are multiplied?
Is it: (the number of ways of selecting 2 that work out of a total of 16 that work) times
(the number of ways of selecting 1 that doesn't work out of a total of 4 that don't work)? . . . . yes
Why multiply?

That's one of the fundamental rules of Counting.

We want: 2 that work and 1 that doesn't work.

A simplfied version:
. . With an "and" statement, we multiply the numbers.
. . With an "or" statement, we add the numbers.

Would the method shown below of doing the questions be acceptable?

\(\displaystyle i)\;\;\frac{16}{20}\cdot\frac{15}{19}\cdot\frac{14}{18} \:=\:\frac{28}{57} \:=\: 0.491\text{ (3 s.f.)}\)

Absolutely!

\(\displaystyle ii)\;\;3\cdot \frac{16}{20}\cdot\frac{15}{19}\cdot\frac{4}{18} \:= \:\frac{8}{19}\:=\:0.421\text{ (3 s.f.)}\)

\(\displaystyle \text{If it was done this way, do you multiply by 3 because of }{3\choose2}\)
or in other words, being 3 ways to arrange them?

Yes . . . That is exactly the reasoning.

\(\displaystyle \text{With: }\:\frac{16}{20}\cdot\frac{15}{19}\cdot\frac{4}{18} \text{, we have established the order: Pass-Pass-Fail.}\)

. . and there are, of course, three possible orders.

Can that method also be used for questions that can be done with binomial distribution?

Yes . . . the Binomial Formula is simply faster.

\(\displaystyle \text{If }P(\text{pass}) \:=\:\frac{4}{5},\;\;P(\text{fail}) \:=\:\frac{1}{5}\)

. . \(\displaystyle P(\text{3 Pass}) \:=\:\frac{4}{5}\cdot\frac{4}{5}\cdot\frac{4}{5} \:=\:\frac{64}{125} \:=\:0.512\)

. . \(\displaystyle P(\text{2 Pass, 1 Fail}) \:=\:{3\choose2}\cdot\frac{4}{5}\cdot\frac{4}{5}\cdot\frac{1}{5} \:=\:\frac{48}{125} \:=\:0.384\)

So a binomial table/formula cannot be used because the events are dependent
(i.e. the probability of the next outcome changes each time one is chosen)? . . . . Right!
Is that what you mean by "finite"?

Since there is a finite number of objects (20 in our example),
. . the probabilities change with each drawing.

With a Binomial problem, the number of objects is assumed to be infinite,
. . so the probabilities remain the same regardless of what is drawn.