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

[blackops23]

Hi just a few questions I need help with.

Q1. A bag contains 2 red balls, 1 black ball, and 1 white ball. Andrew selects one ball from the bag and keeps it hidden. He then selects a second ball, also keeping it hidden.
(i) Draw a tree diagram to show all possible outcomes.

Already done, too lengthy to show here.

(ii)Find the probability that both the selected balls are red.
P(RR) = 1/2 * 1/3 = 1/6

(iii) Find the probability that at least one of the selected balls is red.
P(RR + RB + RW + BR + WR) = 0.5 + BR + WR = 0.5 + 1/6 + 1/6 = 5/6
Is there a quicker way to do this, or is this the fastest method?

(iv) Andrew drops one of the selected balls and we can see that and we can see that the ball is red. What is the probability that the ball that is still hidden is also red?

- This is the one i'm having trouble with.
The actual answer is 1/5, but I don't see how this question is any different from PART (ii), because Andrew only has TWO hidden balls, if they are both red, then P(RR) = 1/6. Unless I'm missing something?

Help on Part (IV) would be greatly appreciated,

thanks

 

[Bored Of Fail 2]

for part ii. you use the complementary event.

p(aT LEAST 1 RED ) = 1- p( no reds ) = 1 - ( 2/4 x 1/3 )

 

[Bored Of Fail 2]

part 4 is conditional probability, I will scan up a picture soon

 

[Drongoski]

Bored Of Fail 2

blackops23 has been one of the most active help-seekers last couple of months. He knows he can get good quality help for nothing because there are so many willing talented and helpful souls on bos.

It is hard recruiting any paying student on bos.

 

[Bored Of Fail 2]

http://img502.imageshack.us/f/probability.jpg/

 

[Bored Of Fail 2]

Bored Of Fail 2

blackops23 has been one of the most active help-seekers last couple of months. He knows he can get good quality help for nothing because there are so many willing talented and helpful souls on bos.

It is hard recruiting any paying student on bos.

lol u saying I should stop, besides there is prob a pretty good chance that he doesnt live near me anyway, also he does 4unit, so if he was after a tutor he would a 4unit one , so I dont think im losing any business lol

 

[Bored Of Fail 2]

I should edit the picture and say "restrict our sample space to those events containing atleast ONE RED "

 

[Bored Of Fail 2]

answer = [ 2/4 x 1/3] / [(2/4 x 1/3) + ( 2/4 x 1/3 ) + ( 2/4 x 1/3) + (1/4 x 2/3) + (1/4 x 2/3) ]

 

[Drongoski]

lol u saying I should stop, besides there is prob a pretty good chance that he doesnt live near me anyway, also he does 4unit, so if he was after a tutor he would a 4unit one , so I dont think im losing any business lol

No. I'm not saying you should stop. I'm just pointing out what has been obvious to you - those who are getting free help here are very unlikely to use any one of us to tutor them.

It's not that they don't go for tuition, Many do: to Talent100, Matrix, Prior, Intuition, Truong, Dr Du, Terry Lee etc or a private tutor.

Provide help because you enjoy it; because you want to help. Hope for nothing in return. If someone decides to engage you off bos, that's an unexpected bonus.

 

[Bored Of Fail 2]

No. I'm not saying you should stop. I'm just pointing out what has been obvious to you - those who are getting free help here are very unlikely to use any one of us to tutor them.

It's not that they don't go for tuition, Many do: to Talent100, Matrix, Prior, Intuition, Truong, Terry Lee etc.

Provide help because you enjoy it; because you want to help. Expect nothing in return. If someone decides to engage you off bos, that's an unexpected bonus.

I understand that. However, if someone on this site did live close to you, and hadnt already started tutoring with another company, and they have been getting alot of help off here from you, and then they go to a local tutoring company instead of atleast trying you out first ( especially if you were doing first hour free and cheap rates ) then I do find that a bit rude.

 

[Drongoski]

Unfortunately that's how the world is. They may think you are good but somehow you are just like one of them; there is a sense of familiarity. You know the saying? "No prophet is acceptable in his own country".

Besides if they want your help, they know they can get it for free on bos.

 

[blackops23]

answer = [ 2/4 x 1/3] / [(2/4 x 1/3) + ( 2/4 x 1/3 ) + ( 2/4 x 1/3) + (1/4 x 2/3) + (1/4 x 2/3) ]

Hey thanks so much, but still don't understand why the sample space must be restricted... :S

But ignoring that, could you please help me on a HSC question which I don't understand the WORDING of.

Q. David has invented a game for 1person. He throws 2 ordinary dice repeatedly until the sum of the two numbers shown is either 7 or 9. If the sum is 7, David LOSES, if it is 9, David WINS. If the sum is anyother number, David throws it again until it is a 7 or 9.

(i) Show the probability that David wins on the first throw of the dice is 1/9.
Easy enough to do.

(ii) Calculate the probability that a second throw is needed.
For David to do another throw, he must not get a 7 or 9.
P(7) = 3/18, and P(9) = 1/9
So P(7 or 9) = 5/18
Complentary events: P(no 7 or 9) = 1 - 5/18 = 13/18

(iii) What is the probability that David wins on his first, second or third throw?
P(win on 1st throw) = 1/9
P(win on 2nd throw) = 13/18 * 1/9
P(win on 3rd throw) = (13/18)^2 * 1/9
Adding these all together: P(E)=727/2916

Now the problem:

(iv) Calculate the probability that david wins the game
P(E) = (13/18)^n * 1/9?? hahaha I got no idea...
Seriously, the question doesn't specify how many throws, so how do I calculate the probability? By the way answer is 2/5.

Thanks so much, appreciate the help!

 

[blackops23]

Unfortunately that's how the world is. They may think you are good but somehow you are just like one of them; there is a sense of familiarity. You know the saying? "No prophet is acceptable in his own country".

Besides if they want your help, they know they can get it for free on bos.

hey hey hey, that's kind of unfair saying that don't you think? I live in Minto by the way, far south-west of sydney, and every guy I see on BOs offering Tutoring services, lives in Bankstown, Parramatta, Epping, Blacktown, and Fairfield, all on the wrong side of Sydney, for me.

I know ppl like Bored Of Fail 2, deterministic and jm01, don't have to help me, but they still give up their time to go out of their way to do so, and you've no idea how much I really appreciate these kind gestures. Yr 12 starts in a few days, so any problems I get with maths, I won't have time to post on BOS, so really would prefer 1 on 1 tutoring, despite the heavy cost, so please stop depicting me as a slimy little knowledge sucking parasite/vermin that unfairly asks for help.

 

[Bored Of Fail 2]

Hey thanks so much, but still don't understand why the sample space must be restricted... :S

because we know some extra information about the events, we know that one of the balls MUST be red, therefore the probability of getting atleast one red has now been "made" 1 ( a certain ). Note however that if we sum up all the probabilties of getting atleast one red on the tree diagram them we will not get "1" .

We are taking a ratio of probabilities, to come up with the probability, compared to the new sample space, the probability of two reds occuring is the ratio of the probabilites of two reds occuring "normally" ( without us knowing there is atleast one red ) to the probability of atleast one red occuring

hopefully that makes some sense , its very tricky to explain

 

[Bored Of Fail 2]

hey hey hey, that's kind of unfair saying that don't you think? I live in Minto by the way, far south-west of sydney, and every guy I see on BOs offering Tutoring services, lives in Bankstown, Parramatta, Epping, Blacktown, and Fairfield, all on the wrong side of Sydney,

I know ppl like Bored Of Fail 2, deterministic and jm01, don't have to help me, but they still give up their time to go out of their way to do so, and you've no idea how much I really appreciate these kind gestures. Yr 12 starts in a few days, so any problems I get with maths, I won't have time to post on BOS, so really would prefer 1 on 1 tutoring, despite the heavy cost, so please stop depicting me as a slimy little knowledge sucking parasite/vermin that unfairly asks for help.

relax man, hes not attacking you , he was just making a general statement

 

[Bored Of Fail 2]

your next question involves infinite sums, nice connection between series and probability

 

[blackops23]

relax man, hes not attacking you , he was just making a general statement

yeah ok ,sorry my bad drongoski

Also regarding the question, I'm not sure if these are 2U or 3U hsc questions, but I'd always been under the impression these were 2U? Unless this is the norm for 2U?

 

[blackops23]

hahah think I got it

Probability that he wins is P(he wins)/P(he wins)+(P he doesn't win)

I used the notion of limiting sums, thanks, I made the number of throws = 10
So P(he wins) = (13/18)^10 * 1/9
P(he doesn't win) + P(he does win) = (13/18)^10 * 3/18 + (13/18)^10 * 1/9

Divide those two together, and you get 2/5

Now if only someone explained the logic to me... Because what I did seems more like a fluke than anything else...

 

[Bored Of Fail 2]

so the probability of him winning is actually an infinite sum, there are an infinite amount of ways he can win

he can win on the first throw, the second, the third, or the millionth

so its just continuing the pattern found in part c

answer = (1/9) +[ (1/9) x (13/18) ] + [ (1/9) x (13/18)^2 ] + ..... infinite sum

now factor out the (1/9)

(1/9) [ 1 + 13/18 + (13/18)^2 + .... ] the part in the brackets is an infinite sum, a=1, r=13/18

ans= (1/9) x a/ (1-r) = (1/9) / [ 1-13/18]

 

[Bored Of Fail 2]

see above post for logic lol