More Help with Past HSC Probability Questions Needed! (1 Viewer)

[ohexploitable]

Oh crap, I got one more question:

Q. The die used in a game has 20 faces. Each face has a different letter of the alphabet marked on it; however the letters Q, U, V, W, X, Y, Z have not been used.
(i) The die is rolled twice. What is the probability that the same letter appears on the upper face twice?

Answer was 1/20, I thought it would be 1/400? Am I wrong, Why?

P(rolling any face)=1
P(rolling the same face again)=1/20

P(E)=1*(1/20)=1/20

 

[Drongoski]

Oh crap, I got one more question:


Answer was 1/20, I thought it would be 1/400? Am I wrong, Why?

You want P(AA or BB or CC or . . . or TT) = P(AA) + P(BB) + . . . + P(TT) = 20 x (1/400) = 1/20

Here AA, BB, CC . . . are all mutually exclusive.

 

[ohexploitable]

hahaha disregard that i suck cocks

 

[Drongoski]

yeah ok ,sorry my bad drongoski

I'm sorry too. If you can find your way to Epping/Eastwood I'd be happy to give you a 2-hr lesson on probability or any new topic (free of course) this week, before school resumes.

 

[blackops23]

Thanks, can't believe I didn't see there were 20 possibilities... Guess it's much like getting a double on a die which is 1/6

 

[ohexploitable]

But it said the same letter TWICE. Say you get one E, roll it again, another E. Isn't 1/20 a little to high of a probability? Please explain.

The dice has 20 sides, you want a specific side (E),

event space = 1
sample space = 20

P(E)=1/20

 

[blackops23]

P(two 'E's)=1-[P(no 'E's)+P(three 'E's)]

=1-[(19/20)³+(1/20)³]
=57/400

?? The answers say 57/8000, and I got 57/8000...??

 

[ohexploitable]

Oh wait... I forgot P(one 'E'). lololol

 

[blackops23]

Oh wait... I forgot P(one 'E'). lololol

So for this Q, do you reckon not taking the complementary and just plain counting would be faster? Because there are only like 3 outcomes.

 

[ohexploitable]

So for this Q, do you reckon not taking the complementary and just plain counting would be faster? Because there are only like 3 outcomes.

lol don't ask me, I haven't actually done this topic yet.

 

[Bored Of Fail 2]

whats so hard about the complementary event?? it is 4 terms, if you try and count out all the possibilities you will be there for ages, the complementary event is a powerful concept that can simplify a lot of probability questions.

Usually the complementary event is the easiest way

 

[Drongoski]

So for this Q, do you reckon not taking the complementary and just plain counting would be faster? Because there are only like 3 outcomes.

Your method already good. P(exactly 2 E's in 3)

= P(EEx or ExE or xEE) = P(EEx)+P(ExE)+P(xEE) = 3 x P(EEx) = 3 x [1/20 x 1/20 x 19/20] = 57/8000

 

[blackops23]

whats so hard about the complementary event?? it is 4 terms, if you try and count out all the possibilities you will be there for ages, the complementary event is a powerful concept that can simplify a lot of probability questions.

Usually the complementary event is the easiest way

I know, but for this one there was only 3 outcomes by counting...

 

[Bored Of Fail 2]

I know, but for this one there was only 3 outcomes by counting...

lol to be honest i didnt really read the question in full detail , but generally the complementary event is easier , in this case they are both about the same , but usually complementary events are the easiest

especially if you see the word "atleast" in a probability question

 

[deterministic]

@blackops : have you done permutations and combinations yet? Because some of these questions can be done quickly using them.

 

[blackops23]

@blackops : have you done permutations and combinations yet? Because some of these questions can be done quickly using them.

nope, my school is pretty damn slow

 

[Lumix]

You should either get a tutor or self-learn. If you have the Cambridge 3u Year 12, you can teach yourself Perms and Combs easily. You can skip the first few exercises as they are pointless in my opinion and just start straight off with Perms and combs. And skip the binomial probability if u havent done binomials yet.