Question

A multiple choice test has 10 questions. Each question has four answer choices.

a. What is the probability a student randomly guesses the answers and gets exactly six questions correct?

b.Is getting exactly 10 questions correct the same probability as getting exactly zero correct?

c. Describe the steps needed to calculate the probability of getting at least six questions correct if the student randomly guesses.

Answer 1

@e.mccormick

Answer 2

OK... have you gone into binomeal expansion?

Answer 3

Yeah a little.

Answer 4

Guess what this will need...

Answer 5

binomial expansion...?

Answer 6

Yah...

Answer 7

So if your sucess ia 1/4 and the falure is 3/4

Answer 8

(p+q)?

Answer 9

Yes, that is what you could set those two. Values like that, then find where your specific case falls, and expand that.

Answer 10

is it ^8?

Answer 11

^8? For what?

Answer 12

(p+q)^8

Answer 13

There are 10 questions, \((p+q)^{10}\)

Answer 14

Oh duh. So can I just use Pascal's triangle to expand it?

Answer 15

Yes, if you have the triangle that deep, or the theorem for binomial expansion.

Answer 16

whats the theorem?

Answer 17

\[(p+q)^n=\sum_{k=0}^n\left( \begin{matrix}n\\k\end{matrix}\right)p^{n-k} q^k\]

Answer 18

Where \[\left( \begin{matrix}n\\k\end{matrix}\right)=\frac{n!}{(n-k)!k!}\]

Answer 19

o.O I am so confused.

Answer 20

For low values of n, Pascal's Triangle is better. But if say n=200 and you want the 123rd term, the formula is better.

Answer 21

okay so it would be (p+q)^10 on the left but what is k in the formula?

Answer 22

It is a start point in the sigma sum.

Answer 23

If you have not seen sigma, \(\Sigma\) notation, stick with Pascal.

Answer 24

I have seen it. would k=1?

Answer 25

It starts at 0 and goes to n.

Answer 26

Know how in any line of pascals, the first term is \(x^n\) and the second is \(x^{n-1}y\)? That is because k starts at 0. \(x^n=x^ny^0\) and \(x^{n-1}y=x^{n-1}y^1\)?

Answer 27

so far I have \[{p+q}^2 =\sum_{k=0}^{10}\] right?

Answer 28

\[(p+q)^{10} =\sum_{k=0}^{10}\]

Answer 29

right. so it should be \[(p+q)^{10}=\sum_{k=0}^{10} (\frac{ 10! }{ (10)!0! }) p ^{10} q ^{0}\]

Answer 30

well, uhhh... that would be... sort of all expanded for the 10th spot. Let me back it up a step.

Answer 31

Okay. sorry.):

Answer 32

\[(p+q)^10=\sum_{k=0}^10\left(\frac{n!}{(n-k)!k!}\right)p^{n-k} q^k\]

Answer 33

Oops!
(p+q)^{10}=\sum_{k=0}^{10}\left(\frac{n!}{(n-k)!k!}\right)p^{n-k} q^k

Answer 34

Aaargh! Forgot my braces. LOL
\[(p+q)^{10}=\sum_{k=0}^{10}\left(\frac{n!}{(n-k)!k!}\right)p^{n-k} q^k\]

Answer 35

When k=0, \(p^{n-k}=p^{10}\), right? That is the spot for all to questions randomly being right.

Answer 36

right.

Answer 37

all to means all ten... don't ask me how it became "to" in my mind. Dyslexia is fun.

OK, the questin asks about when 6 are right. So, when is \(p^{n-k}=p^{6}\)?

Answer 38

when k=4.

Answer 39

Yes! So, we put 4 in there.
(By the way: We are answering a and c at the same time.)

Answer 40

oh sweet! haha so then it should be \[(p+q)^{10}=\sum_{k=4}^{10} (\frac{ 10! }{(6)!4!}) p ^{10}q ^{4}?\]

Answer 41

\[\left[(p+q)^{10}\right]_4=\sum_{k=0}^{10}\left(\frac{10!}{(10-4)!4!}\right)p^{10-4} q^4 \]

Answer 42

Your p also needed to change in the right side... But close.

Answer 43

\[\left[(p+q)^{10}\right]_4=\sum_{k=0}^{10}\left(\frac{10!}{6!4!}\right)p^{6} q^4 \]

Answer 44

okay so how do I evaluate \[[(p+q)^{10}]_{4}\]

Answer 45

Well, the right ahnd side is what that is equal to. We are saying, with the subscript, take the 4th spot. The right hand side becomes that. Now, have you done the wonderful factorials there?

Answer 46

working on it

Answer 47

\[\frac{
10\cdot 9\cdot 8\cdot 7\cdot 6\cdot 5\cdot 4\cdot 3\cdot 2\cdot 1}{
(6\cdot 5\cdot 4\cdot 3\cdot 2\cdot 1)(4\cdot 3\cdot 2\cdot 1)}
\\
\frac{
10\cdot 9\cdot 8\cdot 7}{
(4\cdot 3\cdot 2)}
\\
5\cdot 3\cdot 2\cdot 7=210\]

Answer 48

Yeah I got that too.

Answer 49

Now what?

Answer 50

got a call... sry... This is taking a bit.

Answer 51

No worries.(:

Answer 52

Boss at hardware store needed list of parts... OK, so:
\(210p^6q^4\) is where we stopped.

Answer 53

yes.

Answer 54

p is the chance of being correct in a single event, so \(p=\frac{1}{4}\). Now, what was q?

Answer 55

\[\frac{ 3 }{ 4}\]

Answer 56

So we put those in that formula for that term.

Answer 57

wait so are we done with the sigme part now?

Answer 58

sigma

Answer 59

Yah, we used it to find the proper term. That was the goal of sigma. Instead of finding all the terms, we found just the one we needed.

Answer 60

\[210\left(\frac{1}{4}\right)^6\left(\frac{3}{4}\right)^4\]

Answer 61

alright so p is \[2.44X10^{-4}\]

Answer 62

and q is 0.316

Answer 63

right?

Answer 64

The fractions may be easier to deal with. I tend to leave things in fractional form until the last moment. When you become friends with fractions, the chances of you making an error or bad approximation become less.

Answer 65

okay so p=\[\frac{ 1 }{ 4096 }\]

Answer 66

and q= \[\frac{ 81 }{ 256 }\]

Answer 67

Yes, which is a power of 2 based number... like 256 is!

Answer 68

\[210\left(\frac{1}{4096}\right)\left(\frac{81}{256}\right)\]

Answer 69

\[\frac{210}{1}\left(\frac{1}{4096}\right)\left(\frac{81}{256}\right)\]\[\frac{105\cdot 2}{1}\left(\frac{1}{4096}\right)\left(\frac{81}{256}\right)\]\[\frac{21\cdot 5 \cdot 2}{1}\left(\frac{1}{4096}\right)\left(\frac{81}{256}\right)\]\[\frac{21\cdot 5 }{1}\left(\frac{1}{2048}\right)\left(\frac{81}{256}\right)\]

Answer 70

wait, what happened to the 2?

Answer 71

It ran away with the 4096.

Answer 72

Oh duh!

Answer 73

hehe. Now, because the bottoms are all powers of 2 and the tops no longer have any 2s in them, there is no more similifications.

Answer 74

so I got .016

Answer 75

\[\frac{8505}{524288}\approx 0.0162220001220703125\]

Answer 76

Yep.

Answer 77

So part a is done right?

Answer 78

And c. Well, c is describing what we did.

Answer 79

Depending on if the teacher wants an exact answer, decimal to so many places, or precentage chance, you will need to pick a format for the answer, but that is all there in the numbers.

Answer 80

Okay now what about b. I am so thankful for all your help by the way.

Answer 81

Now, the key to "b.Is getting exactly 10 questions correct the same probability as getting exactly zero correct?" is in those fractions and the rules of the Pascal's Triangle.

What are the first and last parts of pascals triangle, always, 100% of the time, never any other values.

Answer 82

This is also exposed in choose 10 of 10 \(=\frac{10!}{10!}\)

Answer 83

The answer is yes right?

Answer 84

We are not there yet. Hehe. The start and end of Pascal's are always 1, right? Just that far at this point.

Answer 85

okay haha right.

Answer 86

So we have 10 right being \(1p^{10}q^0\) and 0 right being \(1p^{0}q^{10}\)

So really, if \(p=q\) they would be the same.

Answer 87

Does \(p=q\)?

Answer 88

no.

Answer 89

Chance of all 10 right random guessing is:\[\frac{1}{1048576}\]
Chance of all 10 wrong random guessing is:\[\frac{59049}{1048576}\]
So guessing is really, really bad.

Answer 90

Yah, if chances are equal, the curve is pretty evenly shaped. But when chances get higher for one thing or another, the results skew a lot.

Answer 91

So the probability of getting 10 correct is not the same as getting zero correct.

Answer 92

Exactly. No where near the same. In fact, remember our 6/10 thing? where we got 0.0162220001220703125 as the probability? Well, 0/10 is 0.05631351470947265625. That is a bigger number!

Answer 93

You have a better chance of getting none correct than of getting 6!

Answer 94

Hahah holy cow. Thank you so so so so much!(:

Answer 95

Yah, it is almsot 3.5 times more likely that you get 0 right vs. 6. So guessing is really a bad idea.

Answer 96

And it is 59049 times more likely that you get none right vs all 10..... super bad odds there. LOL.

Well, have fun! I think you can go back over this and pull out how the Binomial Theorem and factorials lead to a single chance equation, then use that for the explanation in c.