# Help on Algebra 2 question! :/?

Right my teacher assigned me some homework for the weekend and one of the questions I don’t understand at all! :/ Can you help me figure it out?

If you guess the answers at random, what is the probability of getting at least three correct answers on a five-question true-or-false quiz?

Thanks for your help! :)

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## 28 Answers

I find this one pretty simple; the hard part for me is helping you figure it out while not outright giving you the answer. Here goes…

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The first step is to figure out how many different ways you could answer all of the questions; how many permutations are there? There are two ways to answer each question, and five questions, so you should be able to figure this part out pretty easily.

The second part is figuring out how many of those have at least three correct answers.

Lastly, you divide the first part (total number of possibilities) by the second part (how many have 3+ correct answers) to get your final answer.

So since there are 32 possible answers altogether on the quiz you then divide by ten?

Which then says the possibility of having three correct answers is 10/32 or 0.3125?

I just don’t get it though. Why would you choose 10 at the end? What has the 10 got to do with the three correct answers? I just can’t get my head around it haha.

There is only one way to get all five correct.

There are five different ways to get four correct as you could miss any one (but *only* one) of the five questions, which means that there are *six* ways to get either 4 or 5 correct (5+1=6). Does that make a bit more sense?

But I still don’t get how the 10 is related to the 3 correct answers?

Where is the 10 coming from?

At the end you say 10/32 to get the possibility? Or is that wrong?

At the end, you divide how many ways there are to get at least three questions right by 32 (the total number of possibilities). He never said to divide 10/32. We’re trying to understand where you got 10.

(Number of possibilities that have 3+ correct answers) **/** (Total number of possibilities)

In this case:

(Number of possibilities that have 3+ correct answers) **/** 32

Okay so I understand the part about the total number of possibilities but I just don’t understand how to get how many ways there are to get three correct answers?

I figured out where the 10 came from. Now re-read my second response ;)

You are looking for three *or more* correct answers.

@jerv never mentioned 10, which must mean that you correctly figured that out yourself. Consider all the possibilities for getting 3 right. You could get 1,2 and 3 and the last two wrong, or you could get 2, 3 and 4 right and the other two wrong. C(3,5) is the total number of ways this could happen. Therefore the probability of getting exactly 3 right is C(3,5)/32 = 10/32. To find how many ways of getting at least 3 right, do the same analysis for getting exactly 4 right and for the number of ways of getting all 5 right. Add these together and you get the probability of getting at least 3 right.

Ok I’m beginning to understand a little but what does the C(5,3) mean?

Combinations; picking three three elements out of five possibilities. In this case, choosing three answers out of five.

@LostInParadise I was actually looking at it as more of a bell curve, just like dice, and workig from the ends towards the middle.

Ok so then how does it go from being C(3,5)/32 to being equal to 10/32?

If you pick three elements out of five possibilities, there are actually twenty ways to do so. For instance, if you pick #1 as your first one, you have four numbers to pick for your second choice while if you pick #2 as your first one, you *still* have four numbers to pick for your second choice. Repeat that for all five possible first picks and you get:

4 * 5 = 20

But half of those are “mirrors”; it makes no difference if you choose (1 and 2) or (2 and 1).

20 / 2 = 10

Therefore, there are **ten** different ways you can select exactly three elements out of five.

Or you could just “sum the series”; 4+3+2+1 = 10. By that, I mean;

If you pick #1 then you have four choices for your second pick.

If you choose #2, you have three choices for your second pick that are *not* #1.

If you choose #3, you have two choices for your second pick that are *not* #1 or 2….

Oooh ok. Hahah well sum the series sounds a bit easier! Does it work every time though?

I tend to look at numbers differently than most people.

For a problem like this, yes.

Ok well you were really helpful! More helpful than my teacher could have been! Thanks

No problem!

I was only dealing with one student, and nobody else in the room was causing a disruption while teachers have to deal with far more students at a time, not all of whom behave thus making them part teacher, part zookeeper. They can’t always take the time to be helpful, and it’s not their fault.

I have a web site on combinatorics that you might find helpflul

You can use the Pascal triangle to help solve this and many similar problems. The number of ways to pick X objects out of a set of Y objects is the (X+1)th number on the (Y+1)th row of the Pascal triangle. This means that there are 10 ways to get 3 out of 5 right.

I would also like to point out that @jerv ‘s method of summing the series is flawed (in a way). It always gives the third number on a given row (and the third from the end). If you are asking for the number of ways to three right, you want the fourth number on a row. On the 6th row, the third and fourth number just so happen to be equal, so it works in this case. If there were six questions on the test, there would be 20 ways to get three right, but the sum of the series would be 15.

@PhiNotPi Correct. Summing the series only works for figuring out how many ways to get *exactly* two answers either right or wrong. If you wanted to figure out how many ways to get other numbers right/wrong, that is a different problem and requires a different solution.

Wait a second, I just realized that the question says ”*at least* three correct” instead of “three correct”. This means that you have to add the way to get 3 right and the ways to get 4 right and the ways to get 5 right. That changes it from 10 ways to get three right to 16 ways to get *at least* three right.

Bingo! I actually said that earlier, but…

The lesson here is “Read carefully”.

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