Namu has a Question (1 Viewer)

[lyounamu]

there r 4 multiple choice Q in a test. 4 each Q there is a probability of 1/3 that bob answers it correctly.

Find the probability that the fourth Q Bob attempts is the 2nd that he answers correctly.

 

[lyounamu]

btw this is Shuning not namu..... incase ppl find the title of this thread arrogent and annoying....

 

[kickass91]

edit: 3 x [ (2/3)^2 x (1/3)^2 ]

i meant 4/27

 

[shuning]

回复: Re: Namu has a Question

edit: 3 x [ (2/3)^2 x (1/3)^2 ]

i meant 4/27

me and my friend got this as well.... but Namu said its wrong.... got into a massive arguement so we said, Lets post it on BOS and see how others do it.

 

[addikaye03]

is it 8/27?

i would have thought same as you guys though

 

[lyounamu]

lol shuniing, what a way to "invade" my account and use it for your advantage xD.

From my point of view, we shouldn't assume that the 4th attempt means winning. It can be either winning or losing...well that's from my own interpretation.

 

[study-freak]

I know that i am bad at probability but is it 1/27?
What I did: 3C1(1/3)(2/3)^2 x (1/3) x (1/4)

 

[annabackwards]

This is how i worked it out
Chance of the 4th Q being the 2nd correct
= P(only 1 out of first 3 correct) x p(correct)
= 3C1 x (2/3)^2 x (1x3) x (1/3)
= 4/27

 

[lychnobity]

I know that i am bad at probability but is it 1/27?
What I did: 3C1(1/3)(2/3)^2 x (1/3) x (1/4)

I did something similar, but without the 1/4 ...

3C1 x (1/3) x (2/3)² x (1/3) = 4/27

EDIT: what annabackwards did

 

[shady145]

From my point of view, we shouldn't assume that the 4th attempt means winning. It can be either winning or losing...well that's from my own interpretation.

i think it would mean it could be either winning or loosing if the question didnt say "is the 2nd that he answers correctly."
mabe if it said possibly the 2nd that he answers correctly.
so i did it like this.
if the fourth Q is the second he answers correctly then he has answered only one previous Q correctly... u know the math lol so i took it as it has to be correct. yea 4/27

 

[study-freak]

I put 1/4 because if I have 1 right in 3 Q attempted and have one more correct one, in my arrangement, the additional correct attempt can be placed in any of the places labelled x in: x @ x @ x @ x where @ stands for previous attempts.
But I want the x to be the last one. Hence 1/4

Correct me if I am wrong.

 

[annabackwards]

I put 1/4 because if I have 1 right in 3 Q attempted and have one more correct one, in my arrangement, the additional correct attempt can be placed in any of the places labelled x in: x @ x @ x @ x where @ stands for previous attempts.
But I want the x to be the last one. Hence 1/4

Correct me if I am wrong.

I've never heard of your interpretation but i sort of get your logic...I'm just going to stick to what i did though because probability sucks and i don't want to think about it XD

 

[study-freak]

I've never heard of your interpretation but i sort of get your logic...I'm just going to stick to what i did though because probability sucks and i don't want to think about it XD

Hmm then do you have sets of rules that you memorise for tackling probability questions? I tend to do these questions just by inspection, except for very basic ones like arrangement in a circle is (n-1)!

 

[Miz Janani]

I got 4/27

What was the correct answer?

 

[shady145]

^i dnt think so most peopel got 4/27
it looks like u only did one way that he could have gotten one right before the fourth Q. there are 3 questions in which he could have gotten right so u need 3 cases

 

[runoutofsleep]

wtf you dorks it's 4/27 check out my elegant proof

 

[eriito]

lol what did you write that on?

 

[runoutofsleep]

a car shaped post it note just like fibonacci

 

[annabackwards]

Hmm then do you have sets of rules that you memorise for tackling probability questions? I tend to do these questions just by inspection, except for very basic ones like arrangement in a circle is (n-1)!

Not really a set of rules; i do inspection too but i've never interpreted the way you did XD I seem to identify exactly what sort of method i need to follow after reading a question though ^^

Oh and i have to interpret the circle questions too or i get confused for eg it's (n-1)! because you have to fix one person so there's (n-1)! ways that everyone else can be arranged

 

[ascentyx]

The way i read the question i think he would need a success on the 4th question.

For it to be the second success he would need 1 success in the previous 3 questions.

X ~ Binomial ( 3, 1/3)

So isn't it 3c1 * (1/3)^1 * (2/3)^2 * 1/3

ie 4/27.