If a student randomly guesses at 20 multiple choice questions,?

Question

If the student randomly guesses at 20 multiple choice questions, find the probability that the student gets exactly four correct. Each question has four possible choices. How do you calculate this!!

Answer 1

the way to do this correctly is:

(.25)^4 X (.75)^16=3.91508E-05


HOWEVER YOU AREN'T FINISHED!

you need to take into account all the different possibilites of getting 4 questions right. (i.e. you could get the first four right, or you could get the last four right, or you could get the first two right and the middle two right, etc etc etc)

[20!/(4!*16!)] *3.91508E-05 = 0.189685455

you have an 18.9685% chance of getting 4 questions right if you randomly guess

Answer 2

this is a binomial probability in which there are only two possible outcomes to each event, correct or incorrect. Since there are four choices for each answer, the probability of correct is 1/4 and the probability of incorrect is 3/4. Here is the binomial probability:

C(20,4)*(1/4)^4*(3/4)^16

You need the combination component because out of the 20 questions, there are many ways to get 4 correct: maybe you'll get the first four correct, or maybe it will be the 4th, 7th, 8th, and 19th. Obviously, there are many combinations of 4 when chosen from 20.

Answer 3

This is an application of the binomial distribution. There are C(20, 4) ways of picking 4 out of the 20 questions, which is 20!/(4! 16!), and each possible subset has probability (1/4)^4*(3/4)^16

The probability is 20!/(4! 16!) * (1/4)^4 (3/4)^16

Answer 4

uestions like this are solved using the binomial distribution.

p( quessing 1 question right) = .25
p(getting 1 problem wrong) = .75

4 correct, 16 incorrect

p(exactly 4 correct) = 20C4 * .25^4 * .75^16

Answer 5

C(20,4) . (1/4)^4 . (3/4)^16 = .18968545486586659680

One more interesting thing is getting 5 correct is more
probable ;)
p(5) = .20233115185692436992

p(6) is lesser .16860929321410364160

most probable no. of correct answers is hence 5. :)

Answer 6

Chance of getting one correct is 1/4. So the chance of getting one wrong is 3/4

So... To get exactly 4 correct...

(1/4)^4 * (3/4)^16
**************************
Oops, yeah, I forgot the "choice" part...

20! / (16!*4!) * (1/4)^4 * (3/4)*16

Answer 7

The Binomial distribution has the prospect function P(x=r)= nCr p^r (a million-p)^(n-r) r=0,a million,2,.....,n the place nCr = n! / r! (n-r)! n=20 p=a million/4 r=4 (20C4)(a million/4)^4(3/4)^sixteen =0.1897 you will desire a Binomial danger table.

Answer 8

I thought it would be (20!)/(4!*16!) * (.25)^4 * (.75)^16.....