# Math Help - Of all the articles in a box 80% are satisfactory and 20% are not

1. ## 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?

2. This is a binomial distribution;
So let Y=number of satisfactory items, with n=8
$P(Y=5)=8C5\cdot (.8)^5 \cdot (.2)^3 = 0.1468$

3. 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.

4. Originally Posted by Robb
This is a binomial distribution;
So let Y=number of satisfactory items, with n=8
$P(Y=5)=8C5\cdot (.8)^5 \cdot (.2)^3 = 0.1468$
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?

5. 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

$(6C3.3C2)/(9C5)$

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

6. 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

$(6C3.3C2)/(9C5)$

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.

7. 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 $8C5=56$
And each one of those outcomes listed above has probability $(0.8)^5 \cdot (0.2)^3$

8. 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....