just some q's (1 Viewer)

[Pace_T]

Hey all
just wondering if you guys can solve these q's for me please

1. A set of cards consists of a red 1, 2, 3, a white 1, 2, 3, and 4, and a blue, 1, 2, 3, 4, 5. If four of them are selected at random, find the probability that they are all either "1's", "2's" or "3's"


the answer is 1/33


2. Two different numbers are chosen at random from the set 1, 2, 3, 4, 5
what is the probability that they have an even product
the answer is 7/10

3. There are 5 letters, A, B, C, D, E
what is the probability that both B and D have other letters on either side of them?
answer is 1/10

don't u just love probability?
thanks for ur help

 

[munkaii]

Heres question two:

Hypothetically, if you picked an even number first, your second number odd or even will produce even result.

If your first number was odd, you have a 2/4 chance of picking a second card to be even (because you've chosen one card, you are left with two even and 2 odds).

Combining that.

P(even product) = P (even x odd) + P (odd x even)
= 2/5 x 4/4 + 3/5 x 2/4
= 8/20 + 6/20
= 14/20 = 7/10.

Question 3 seems a little vague.
Question 1... 3 decks..pick four cards? How can you pick 4 1's?

 

[Pace_T]

i have no idea about 1 and 3 because i dont understand the q's lol
i think my hingrish is preaty bad

 

[acmilan]

The answer for 1 would be 0?

 

[acmilan]

For 3: The only way that they can either have letters on either side if they are in the form of:

(x B x D x) or (x D x B x) (where x can be A C or E in any order)

The total number of arrangements possible is 5! (if B and D are allowed to move)
If B and D remain stationary, then the possible number of arrangements is 3!

3!/5! = 1/20

B and D can also interchange, so multiply the answer by 2

.: Probability = 1/10

 

[Trev]

Hmm... I just made a fool of myself
I should read the question properly - 'other letters', I was wondering why B and D weren't allowed next to eachother

 

[thunderdax]

I think the first question is written up properly, its impossible that would happen as its written. If it was 3 cards chosen it would be
prob=3/¹²C₃
=3/220

 

[Pace_T]

hye thanks for ur replies, greatly appreciated
sorry, but i have some more questions if u don't mind me asking

Prove that for al positive integers n,
1^3 + 2^3 + ... + (n-1)^3 + n^3 + (n-1)^3 + ... + 2^3 + 1^3 = (n/3)(2n^2+1)

1^2 + 2^2 + ... + (n-1)^2 + n^2 + (n-1)^2 + ... + 2^2 + 1^2 = (n^2/2)(n^2 +1)

I am finding it really difficult to prove they are true for n=k+1. Maybe I am doing somehting wrong?

Thanks for any help people

 

[richz]

i think u use the ax^2 +bx +c formula to find the last term

eg. for the first one 1^3 + 2^3 + ... + (n-1)^3 + n^3 + (n-1)^3 + ... + 2^3 + 1^3 = (n/3)(2n^2+1)

let the first term be x=1 so a +b +c = 1 next one would be 4a+2b+c = 8 etc. Then u solve simultaneously and u'll find the last term. Work from there..

Hope that helps

 

[Pace_T]

im sorry i don't understand

could you please explain a little more?
thanks for ur help

 

[munkaii]

i think u use the ax^2 +bx +c formula to find the last term

eg. for the first one 1^3 + 2^3 + ... + (n-1)^3 + n^3 + (n-1)^3 + ... + 2^3 + 1^3 = (n/3)(2n^2+1)

let the first term be x=1 so a +b +c = 1 next one would be 4a+2b+c = 8 etc. Then u solve simultaneously and u'll find the last term. Work from there..

Hope that helps

i think youve got the wrong concept.
I have just begin a small part of mathematical induction @ tutor.
Basically its prove for n=1, n=k, n=k+1

 

[LaCe]

I tried this and it is a bit tricky for proving for the (k+1)th term

1^3 + 2^3 + ... + (k-1)^3 + k^3 + (k-1)^3 + ... + 2^3 + 1^3 = (k/3)(2k^2+1)

for the (k+1)th term

1^3 + 2^3 + ... + k^3 + (k+1)^3 + k^3 + ... + 2^3 + 1^3 = (k+1)/3)(2(k+1)^2+1)

using the assumption for kth term:
(k/3)(2k^2+1) +(k+1)^3 + k^3 = (k+1)/3)(2(k+1)^2+1)
expand and simplify (too much to write)
but if it helps, u can expand the RHS and see what u need for the LHS, but all u need to do is expand the LHS and take out (k+1)/3 and ur there

 

[richz]

i think youve got the wrong concept.
I have just begin a small part of mathematical induction @ tutor.
Basically its prove for n=1, n=k, n=k+1

errr........... i think i know that but there is no last term there, they have given u a sequence with no last term so u need to find the last term.....

and that way u can find the last term.....

if i had the time i would do it for him.... but i need to study for my half yearlys

 

[LaCe]

I have the answer to question 3 of the probability part
3) the question should state that they are arranged in a line
if B and D have other letters beside them then B and D must be in poistion 2, and 4 of the line
arrange other letter 3! ways, multiply by 2 because B can occupy position 4 or 2
so we get 3!*2/5!
=1/10
2) is easy

but 1) does not work
the actual probability comes out as 14/55
as we get 9/12*8/11*7/10*6/9=14/55

 

[richz]

sorry pace i'll explain it after my half yearlies