Thread: Of all the articles in a box 80% are satisfactory and 20% are not

Of all the articles in a box 80% are satisfactory and 20% are not. The probability of obtaining exactly 5 good items out of 8 randomly selected articles is?

0.800
0.003
0.147
0.132
0.013

I have no idea how to work this out. Anyone?

This is a binomial distribution;
So let Y=number of satisfactory items, with n=8

Thanks Robb. But I'm afraid I still don't understand why you did what you did. I'm actually self studying from a SAT 2 book. This sum was actually the last sum under the probability section and the binomial was the previous lesson. I went through the notes and examples of both lessons but this problem was a bit different. I shall be grateful if you can explain it a bit more.

Originally Posted by Robb

This is a binomial distribution;
So let Y=number of satisfactory items, with n=8

This will only be true if there is a large number of items in the box so that the probability of success (picking a satisfactory item) stays more or less the same for the 8 trials.

@OP: What don't you get? You were told it's a binomial distribution problem with n = 8 and p = 0.8. Pr(Y = 5) is required. Where is the trouble here?

The problem is - I don't see how the binomial distribution fits into this.

In the Binomial lesson there were only questions like - whats the 4th term of the distribution, write the 1st three terms of (a+b)^0.5 etc etc

And in probability, what I encountered, that used binomial(although I was just thinking of it as combinations) was questions like - A hotel has 5 single rooms. 6 men and 3 women applied for them. Whats the probability that exactly 3 men and 2 women will be able to rent them?

I didn't know how to do it initially but I learned that it was solved like this -

\left(begin{array} 6&3\ end{array})\right

darn I can't get that format in latex. But I believe its the same as




Concerning our current question, I can't figure out whats being done

Originally Posted by Utterconfusion

The problem is - I don't see how the binomial distribution fits into this.

In the Binomial lesson there were only questions like - whats the 4th term of the distribution, write the 1st three terms of (a+b)^0.5 etc etc

And in probability, what I encountered, that used binomial(although I was just thinking of it as combinations) was questions like - A hotel has 5 single rooms. 6 men and 3 women applied for them. Whats the probability that exactly 3 men and 2 women will be able to rent them?

I didn't know how to do it initially but I learned that it was solved like this -

\left(begin{array} 6&3\ end{array})\right

darn I can't get that format in latex. But I believe its the same as




Concerning our current question, I can't figure out whats being done

If you choose 8 things with replacement (when there are a large number of items you can make this simplification) then the number of ways of choosing exactly 5 good things is 8C5. And the probability of each of these combinations is (0.8)^5(0.2)^3. By the way, the binomial distribution is not the same as the binomial theorem. It arises from a combinatorial argument similar to the one I've given.

Yea, i figured that it was implied in the question that removing an article from the box wont effect the chance of success with the other items... (Either by replacement, or a large number of items in the box)


The chance of an item that is selected from the box being satisfactory is constant at 0.8, and you want to select 8 items in total. You want to calculate the probability that of the 8 items selected, 5 will be satisfactory (with probability 0.8) and 3 will be unsatisfactory (with probability 0.2).

Since the order doesn't matter, you can pull the articles out in any combination that yields 5 satisfactory, and 3 unsatisfactory, ie (letting S=satisfactory, U=unsatisfactory);
SSSSSUUU
SSSSUUUS
SSSSUUSU
SSSSUSUU
.....
UUUSSSSS
So the number of ways that can be done is the
And each one of those outcomes listed above has probability

Hmm something new to me but I think I got it now. So what you do for Q's like this is write the possible number of combinations of obtaining the particular result and multiply that by the probability^(# of desired occurrences) of each event encountered right....