Probability Help PLEASE!! (2 Viewers)

[GaDaMIt]

Number of ways with the T's together and E's together = 4! = 24
Total number of unrestricted ways = 6!/(2!x2!) = 180
.'. probability = 24/180 = 2/15.

Note: I am assuming that the pairs of letters are counted as the same when swapped around.

yeah thats right now.. but i dont get why 24 is the top number on the fraction (LOL forgotten the name of the top number on a fraction =\)

 

[SoulSearcher]

yeah thats right now.. but i dont get why 24 is the top number on the fraction (LOL forgotten the name of the top number on a fraction =\)

Numerator

24 is the numerator because you count the T's and the E's as a single group each. Since there are 4 separate groups, the amount of times they can be arranged in a row is 4! times, or 24 times.

 

[pLuvia]

13) The letters of KETTLE are arranged randomly in a row. Find the probability that:

c) the two letters E are together and the two letters T are together

Treat double E and double T as one unit therefore number of arrangements is 4!
Total number of arrangements 6!/(2!*2!)
Probability - 4!/[6!/(2!*2!)]=2/15

Edit: lol, didn't realise the 4th page

 

[GaDaMIt]

for that same question.. probability that

d) the Es and Ts are together in one group?

 

[Riviet]

If the T's and E's are together in 1 group, we have the K and L left, which makes 3 groups. Arranging these, as well as considering that the 2 T's and 2 E's can be arranged in 4!/(2!x2!) ways:
desired number of ways = 3! x 4!/(2!x2!) = 36
Now total number of ways = 6!/(2!x2!) = 180
.'. probability = 36/180 = 1/5.

 

[pLuvia]

Arrangement of Es and Ts - 4!/(2!*2!)
Arrangement - [4!/(2!*2!)] * 3!
Total number of arrangements - [6!/(2!*2!)]
Probability - [4!/(2!*2!)] * 3!/[6!/(2!*2!)]=1/5

Edit: All three of us

 

[Riviet]

Yay, looks like our answers agree with each other [for once]. Nice job guys.

 

[GaDaMIt]

3! (treating the four letters as a group) x 4! (arrangements within the four letters).

Total number of arrangements is 6!

So 144/720
=1/5

Equally, you could do 3! x (4!/2!2!)/6!/2!2!) but that is exactly the same and I think a bit misleading - surely you should count all repetitions as distinct possibilities when you are doing probability.

I don't get what you just said there? I do get the answer so thats the important bit out of the way

You mean cause the denominator within the numerator and denominator will both be the same it should not be counted or something?

 

[SoulSearcher]

Let's show it this way: (x/y) / (z/y) = x/y * y/z = xy/zy = x/z, as the ys cancel out. webby has just eliminated the denominator from both the fractions.

 

[Riviet]

Basically, you can do probability, taking into account distinct ways or not. If you do take into account the distinct, then you need to be consistent with your working, ie work out number of desirable and distinct ways (numerator) AND work out the total number of distinct ways (denominator). I personally find counting the distinct ways more convincing, as you're actually working it out fully.

 

[SoulSearcher]

You'll probably have to do it the distinct way in the HSC exam to get full marks.

 

[GaDaMIt]

Ahh geez stuck on the next question =\

A tank contains 20 tagged fish and 80 untagged fish. On each day, four fish are selected at random, and after noting whether they are tagged or untagged, they are returned to the tank. Answer the following questions, correct to three significant figures.

c) Calculate the probability of selecting no tagged fish on every day for a week.. probability for one day = 0.403 .. i dont get what do with the "week"

 

[pLuvia]

One day - [²⁰C₀*⁸⁰C₄]/¹⁰⁰C₄

Each day will be the same probability, assuming its 7 days, so it's just
[[²⁰C₀*⁸⁰C₄]/¹⁰⁰C₄]⁷=0.00174

 

[webby234]

You are selecting 28 fish in total. So the answer is 80C28/100C28

= 5.82 x 10^-4

EDIT: oops missed the replacement part - pLuvia is correct.

And I actually think my method is more in full - each repetition is still a possibility - even if the ones you don't count cancel out in the end. What do you think of the example I gave? It would certainly be impossible for markers to mark me down for using my method - it is a perfectly valid way of approaching the question.

 

[GaDaMIt]

Last question for the night.. follows the other one

What is the probability of selecting no tagged fish on exactly three of the seven days during the week

 

[webby234]

This is a binomial question that you may not have covered yet, however I will answer it anyway.

Probability of no tag is 0.403
Probability of at least one tag is 0.597

So probability none on three of the days = (0.403)³ x (0.597)⁴ (chance of getting a tag on four days) times 7C3 (the number of ways to get three of the days).

= 0.291

 

[GaDaMIt]

I'm back

Six people are to be divided into two groups, each with at least one person. Find the probability that:

a) there will be three in each group
b) there will be two in one group and four in the other
c) there will be one group of five and an individual


^^ I don't get what you're supposed to do when its split into two groups..

 

[webby234]

So you can have a group of 1 and group of 5, 2 & 4 or 3 & 3?

This means the total number is (6C1 + 6C2 + 6C3 + 6C4 + 6C5)/2 - it is divided by two as each group of two corresponds to a group of four, each group of one corresponds to a group of five and each group of three is repeated.

So total number is 31

a) 6C3/2 = 10 of these will have three people in them so
10/31

b) (6C2 + 6C4)/2 = 6C2 have two in one and four in the other = 15/31

c) Is the rest of the groups so 6/31

 

[GaDaMIt]

The digits 1,2,3,4 are used to form numbers that may have 1,2,3,4 digits in them. If one of the numbers is selected at random, find the probability that:

d) it is odd and greater than 3000
e) it is divisible by 3

 

[webby234]

d) Total possibilities
= 4P1 + 4P2 + 4P3 + 4P4
= 64

Only four digit numbers can be greater than 3000. First consider numbers begining with 3 (if you try to do them all together you run into problems with the odd part).

3! = 6 4 digit numbers will begin with 3. Now either list them or consider that they will end in 1 (the only possibility for odd) 1/3 of the time. So two of the numbers between 3000 and 4000 are odd

Greater than 4000 - 2/3 of the numbers will be odd (can end in 1 or 3). So four of the numbers will be odd.

Oops - didn't really need to split up the 4000s and the 3000s, but better to be safe.

6/64
= 3/32