i hate probability (1 Viewer)

[1234567]

q1. a box contains 9 marbles of which 4 are red and 5 are blue, three marbles are drawn one a a time and without replacement. find the probability that 3 marbles of each colour are left in the box..

q2. a die with faces numbered 1,2,3,4,5,6 is biased, so that the probability of a score x is given by p(x) = kx, where x = 1,2.....6
find k

 

[McLake]

Q1:

IE FIND the prob that you draw 1 red and 2 blue.

P(RBB) + P(BBR) + P(BRB)
= [4/9 * 5/8 * 4/7] + [5/9 * 4/8 * 4/7] + [5/9 * 4/8 * 4/7]
= 10/21

IS THIS RIGHT???


Q2:

?????????

 

[spice girl]

McLake: your question 1 is right

Q2: P(1) + P(2) + P(3) + P(4) + P(5) + P(6) = 1

k(1 + 2 + 3 + 4 + 5 + 6) = 1

k = 1/21

 

[McLake]

Originally posted by spice girl
McLake: your question 1 is right

Q2: P(1) + P(2) + P(3) + P(4) + P(5) + P(6) = 1

k(1 + 2 + 3 + 4 + 5 + 6) = 1

k = 1/21

Thank you spice girl.

Um, with Q2

I konw that must be right, and it looks right ...

BUT

Why isn't it P(1) + P(2) + P(3) + P(4) + P(5) + P(6) = 21k ?

[p(1) = k, p(2) = 2k ....]

 

[BlackJack]

Q1: yes it's correct.

Q2: Spice girl is right. It's the same thing. It IS 21k, but 21k =1 because it's all outcomes.
Therefore k=1/21

 

[p00_p00]

Originally posted by 1234567
q1. a box contains 9 marbles of which 4 are red and 5 are blue, three marbles are drawn one a a time and without replacement. find the probability that 3 marbles of each colour are left in the box..

q2. a die with faces numbered 1,2,3,4,5,6 is biased, so that the probability of a score x is given by p(x) = kx, where x = 1,2.....6
find k

i got 10/21

 

[p00_p00]

WTF? Q2???


Can some1 explain plz?

 

[BlackJack]

The chances that you'll get each face is different.
p(x)=kx means that to get face of '3', x=3 and the probability fo that face coming up is 3k
Hence, probabiliy to get each face:
p(1)=k
p(2)=2k
p(3)=3k
p(4)=4k
p(5)=5k
p(6)=6k

Also, the sum of all possiibilities is 1... etc...

 

[1234567]

yeah question 2 is 21
question 1 is 10/21
but i dunt quiet understand how u guys did these 2 questions.

 

[McLake]

Originally posted by BlackJack
Q1: yes it's correct.

Q2: Spice girl is right. It's the same thing. It IS 21k, but 21k =1 because it's all outcomes.
Therefore k=1/21

I'm soooooooooooo stupid!!!

Got it now (why I couldn't make the connection, I don't know)

OK, heres Q2 in full:

P(x) = kx
so P(1) = k, P(2) = 2k , etc.

so P(1) + P(2) + P(3) + P(4) + P(5) + P(6)
= (1 + 2 + 3 + 4 + 5 + 6)k
= 21k

but "P(1) + P(2) + P(3) + P(4) + P(5) + P(6)" is equal to all outcomes.

so 21k = 1
k = 1/21

 

[Weisy]

Originally posted by 1234567
yeah question 2 is 21

I got k=1/21 as well...

?

 

[blitz]

Can u do Q1 using binomial probabilities???

 

[spice girl]

My way of doin Q!:

Like McLake said, you draw one red from a possible 4, and 2 blue from a possible 5. Order isn't important (it's a selection).

Thus # possible ways is (4C1)*(5C2) = 4*10 = 40
# sample space = 9C3 = 84

40/84 = 10/21