More Help with Past HSC Probability Questions Needed! (3 Viewers)

[Bored Of Fail 2]

yes what you did would probably get 1/2marks if you are lucky, you cannot assume stuff

 

[blackops23]

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]

But that seems much more easier, thank you very much. =)

Sorry about this but could you please help me on this one as well? I just can't figure how to APPLY probability on this question:

Q. 75 tagged fish were released into a damn known to contain fish. Later a sample of 42 fish was netted from this dam and then released. Of these 42 fish it was noted that 5 fish were tagged (from the original 75). Estimate the total no. of fish in the dam.

Now I do Bio at school, and this is an example of a thing called the capture mark recapture experiment, where you estimate the population of a species in an area based on the proportion of retagged fish blah blah blah. The formula estimating the population is : (75*42)/5 which = 630, which is what the answer is.

My question is, how do I apply probability in the question, i.e. in relation to SAMPLE SPACE and EVENT SPACE. Sample space is n, event space is 75, and 42 & 5, well I don't know =(

Thank you.

 

[blackops23]

sorry for double post

 

[Bored Of Fail 2]

:

Now I do Bio at school.

yes I remember yr11 biology XD, ill have a look at it

 

[Bored Of Fail 2]

it seems more like ratios than probability

so it said that out of 42 fish there were 5 that were tagged

so therefore we would expect 1 tagged fish for every 42/5 fish in the dam ( yes, I know numbers of fish are usually whole numbers )

now we put 75 tagged fish in the dam , therefore there are (42/5) x 75 fish in the dam ( our estimate is anyway )

 

[ohexploitable]

P(tagged fish)=75/n
P(tagged fish)=5/42

75/n=5/42
n=(75*42)/5
n=630

 

[ohexploitable]

But if I ever saw this question, I'd just use ratios.

 

[Bored Of Fail 2]

P(tagged fish)=75/n
P(tagged fish)=5/42

75/n=5/42
n=(75*42)/5
n=630

yes thats a better way to structure the answer

 

[deterministic]

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

Sample space is reduced due to the fact it is a conditional probability (condition is that one of the ball is red). Since you already know one of the ball is red, the sample space must be reduced to the set of outcomes where at least one of the ball is red.

 

[blackops23]

simple as that hmm probably was just over complicating it... xD

Ok, sorry about this but I kinda need a bit more help on a few more Q's. They seem easy at first, but then when I delve into them, my mind just blanks and I can't figure how to solve them. Like this one for e.g.

Q. Pat and Chris each throw a die.
(i) Find the probability that they each throw the same number.
P(E) = 1/6

(ii) Find the probability that the number thrown by Chris is greater than the number thrown by Pat.
Now I could physically count each of the outcomes in which this happens, but I was just wondering,

is there a quicker, more efficient method?

thanks

 

[Bored Of Fail 2]

you have to count them, but it shouldnt take too long to do

 

[ohexploitable]

There's 1/6 chance that they get the same number so there's a 5/6 chance they get different numbers i.e. one will be bigger than the other. Chris will be bigger half of the time.

5/6*1/2=5/12

 

[deterministic]

(ii) Easier way:
3 possible outcomes: They throw the same, Chris>Pat, Pat>Chris
P(not throw the same number)=1-1/6 =5/6
P(Chris>Pat)=P(Pat>Chris) by symmetry
So P(Chris>Pat)=5/12

 

[Bored Of Fail 2]

There's 1/6 chance that they get the same number so there's a 5/6 chance they get different numbers i.e. one will be bigger than the other. Chris will be bigger half of the time.

5/6*1/2=5/12

hmm thats a good way

 

[blackops23]

Thanks guys, promise there's only one more question:

A box contains 12 chocolates all of the same appearance and size. Four chocolates are hard, 8 are soft, Kim eats THREE chocolates chosen randomly from the box.
Find the probability that,

(i) first chocolate Kim eats is hard
P(H) = 1/3

(ii) Kim eats three hard chocolates
P(3 hard chocalates) = 1/3 * 3/11 * 2/10 = 1/55

(iii) Kim eats EXACTLY ONE HARD CHOCOLATE
P(HSS) + P(SHS) + P(SSH) = 1/3 * 8/11 * 7/10 + 2/3 * 4/11 * 7/10 + 2/3 * 7/11 * 4/10
= 28/55

My question: Is there any quicker method than this? (I know the above method isn't that long, but I'm just wondering if there are any ninja methods)

Thanks

 

[Bored Of Fail 2]

for prob exactly one hard

P(HSS) = 4/12 x 8/11 x 7/10
P(SHS) = 8/12 x 4/11 x 7/10
P(SSH) = 8/12 x 7/11 x 4/10

now these are all the same, note the same factors on the top and the bottom, this is why you dont simplify fractions in probability unitl right at the end

so the actual simplified answer is 3 x P(HSS)

this is because there one general way of picking our case, but we can rearrange the order in 3!/2! ways (ie 3ways)

( there are three objects, but two repeated elements )

you will learn bout it in comb/perms

 

[blackops23]

ok thanks, man I can't believe I'm having so much trouble with 2U... =(, but I guess I did only start probability 3 days ago, anyways, thanks guys for all the help time to move on to english =( =( =( =(

 

[Bored Of Fail 2]

ok thanks, man I can't believe I'm having so much trouble with 2U... =(, but I guess I did only start probability 3 days ago, anyways, thanks guys for all the help time to move on to english =( =( =( =(

yeh dont worry bout it, if you havent done it in class yet, and you have only done 3days by yourself then you are doing pretty good, you will understand it alot better once you go over it in class, but you are reaosnably good for someone thats only done 3days teaching themself

 

[blackops23]

yeh dont worry bout it, if you havent done it in class yet, and you have only done 3days by yourself then you are doing pretty good, you will understand it alot better once you go over it in class, but you are reaosnably good for someone thats only done 3days teaching themself

haha thanks, but my teacher told my class to work on the probability topic during the holidays I left it till the end, because of 4U and youtube, lol whats gonna happen if she skips right over it - i'd be screwed...xD

 

[blackops23]

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?

(ii) The die is rolled three times. What is the probability that the letter E appears on the upper face exactly twice?
For this one I drew a PARTIAL probability tree, with 2 OUTCOMES on each BRANCH - P(E) and P(NoE). And then I got the outcomes P(EE NoE), P(E NoE E), P(NoE EE), and then I added these up to get 57/8000

Is there a quicker way?

thanks guys