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? Explain.

c. Describe the steps needed to calculate the probability of getting at least six questions correct if the student randomly guesses. You do not need to calculate this probability!

It would be greatly appreciated if you could show work as to how to solve these, thanks!

Answer 1

@amistre64 @ash2326 @hartnn @Luis_Rivera @.Sam. @mathstudent55

Answer 2

@hba @RadEn @ajprincess

Answer 3

Sweet, thanks, I'm gonna see if I can figure out the other two.

Answer 4

When computing a binomial probability, it is necessary to calculate and multiply three separate factors:

1. the number of ways to select exactly r successes,

2. the probability of success (p) raised to the r power,

3. the probability of failure (q) raised to the (n - r) power.

Answer 5

IDk, I'm still really confused as to what to do, been trying to do this problem for last 30 mins I just don't know how. @kropot72 could you possibly help?

Answer 6

@Mertsj

Answer 7

@campbell_st @Luis_Rivera could you please guide me through the steps more? Been stuck on this for hours :(

Answer 8

@hba @waterineyes

Answer 9

well do you understand binomial probability, where there is a probability for success and a probability for failure...?

Answer 10

Kind of, but not so much, I'm reallllyyy bad at statistics, anything else I can do, just not statistics so much. This is like my last problem could you pleassssseee guide me through? @campbell_st

Answer 11

in a multiple choice question the probability of of success is

\[p = \frac{1}{4}\]

and the probability of failure is

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

does that make sense...?

Answer 12

yes, got that part

Answer 13

ok... so you are looking at a binomial expansion

(p+ q)^10

which could be expanded using pascals triangle....

Answer 14

here, I used mathway to do it faster :P

Answer 15

but since there would be 2 many terms you can use

\[^nC_{r} (p)^{n - r}(q)^r\]

the C stands for combinations.... or how many ways ca you arrange n items in r ways.

Answer 16

so in your question you are given n, p, q which can be used for a specific cases, with different values of r...

\[^{10}C_{r}(\frac{1}{4})^{10 - r}(\frac{3}{4})^4\]

Answer 17

so yes you're correct \[^{10}C_{6} = 210\]

Answer 18

\[^nC_{r}\] just gives the coefficient of the term in that position

so for

(a) you need to evaluate

\[P(6 correct) = 210 \times (\frac{1}{4})^{10 - 6} \times (\frac{3}{4})^6\]

Answer 19

ok so 210x(1/256)x(729/4096)
or
.1459980011?

Answer 20

and hopefully it makes some sense that the correct answers the smaller the probability

for (b)

\[P(10 correct) = 1 \times (\frac{1}{4})^{10 - 0} \times (\frac{3}{4})^0\]

and

\[P(0 correct) = 1 \times (\frac{1}{4})^{10 -10} \times (\frac{3}{4})^{10}\]

Answer 21

for part (c) a least 6 correct..

means you can get 6 correct, 7 correct 8 correct, 9 correct and 10 correct.

So you need to find the probability of each outcome then add them.

\[P(\ge6) = P(6) + P(7) + P(8) + P(9) + P(10)\]

hope this makes sense.

Answer 22

Wait so what I did above was right or wrong?

Answer 23

you answer to part (a) looks correct

Answer 24

Ok, now part b?

Answer 25

How would I figure that out?

Answer 26

so for 10 correct of 0 correct you need to look at the binomial expansion you posted, its just the 1st term (10 correct) and last term (0 correct) to get the answers.

the more questions attempted the greater the chance of not answering them all correct...

Answer 27

have a look at the solution above.

Answer 28

Which one?

Answer 29

Ohh, I didnt see you posted for b

Answer 30

Gimme a sec, i'm gonna try to solve them so you could check my answers?

Answer 31

So for 10 correct I got 9.536743164 e -7
0 correct I got .0563135147, is that right?

Answer 32

yep... they look good...

Answer 33

and part (c) is explained above...

Answer 34

So does that mean that it is not the same probability to get 10 correct and 0 correct?

Answer 35

thats correct... they are different numbers... and it makes sense you have a greater chance of getting a question wrong... or 3 chances in 4... where as a correct answer is 1 chance in 4.

Answer 36

Ok, I'm gonna give c a go, if you could also check my answer when I'm done it would be greatly appreciated.

Answer 37

well the question seems to be just asking...describe the steps.... it doesn't ask for a numerical answer...

so go back up and see how to do part (c)

Answer 38

Oh wheeww, I guess I should read the questions better then xD. Wow I really really really appreciate it, after hours someone finally helps me , thank you soooooooooo much!

Answer 39

hope it makes sense... good luck

Answer 40

Sorry, last question, for these answer, like answer for a for example .1459980011, do I leave that as is as a decimal or am I supposed to like turn it into a percentage or something?

Answer 41

@campbell_st

Answer 42

well its easier to write it as a fraction... avoids confusion.

so for (a) \[\frac{210 \times 729}{256 \times 4096} = \frac {76545}{524288}\]