Probability Q - Help please!

[CSR]

Cant' seem to figure this one out. My friend asked for some help and its bugging me.

Q1) A box has 8 pairs of shoes. 4 Shoes are randomly selected, what is the probability that there will be exactly one complete pair?

Q2) From a team of 6 men and 9 women, what is the probability that 3 men and 2 women are chosen? (Choose 5)

Thanks, so I can go to bed.

[Hellfire]

Cant' seem to figure this one out. My friend asked for some help and its bugging me.

Q1) A box has 8 pairs of shoes. 4 socks are randomly selected, what is the probability that there will be exactly one complete pair?



0% since the box had 8 pairs of shoes, not socks. :cheesygri

[CSR]

0% since the box had 8 pairs of shoes, not socks. :cheesygri

Sorry, typo, shoes only.

[Jay Hova]

Q1) A box has 8 pairs of shoes. 4 Shoes are randomly selected, what is the probability that there will be exactly one complete pair?

(1st, 2nd) + (1st, 3rd) + (1st, 4th) + (2nd, 3rd) + (2nd, 4th) + (3rd, 4th)
(1/8*1/7) + (1/8*1/6) + (1/8*1/5) + (1/7*1/6) + (1/7*1/5)+(1/6*1/5)

?

[mlc2000]

Whats the probability that after you graduate, you will wonder why you studied probabilities instead of a skill that would get you a job?

[EmperorOfCanada]

Cant' seem to figure this one out. My friend asked for some help and its bugging me.

Q1) A box has 8 pairs of shoes. 4 Shoes are randomly selected, what is the probability that there will be exactly one complete pair?

Q2) From a team of 6 men and 9 women, what is the probability that 3 men and 2 women are chosen? (Choose 5)

Thanks, so I can go to bed.

Q1
1 * (1/15) * (1) * (12/13) = 12/195 or 1 in 16.25

Odds of exactly one completed pair? I dont know for sure Im not very good at math :p

(First shoe can be anything) * (one shoe out of 15 remaining makes the pair) * (third shoe can be any of the remaining 14) * (shoe can be any of the 13 remaining EXCEPT the one that completes the second pair)

Q2
24 / 1001

Probably wrong on Q2

[Snicla]

The male and female should be

(3/15) * (2/15) * (1/15) = Answer

But I was never good at this.

[Jon Lai]

Use combinatorics.

[Jay Hova]

Q1) A box has 8 pairs of shoes. 4 Shoes are randomly selected, what is the probability that there will be exactly one complete pair?

(1st, 2nd) + (1st, 3rd) + (1st, 4th) + (2nd, 3rd) + (2nd, 4th) + (3rd, 4th)
(1/8*1/7) + (1/8*1/6) + (1/8*1/5) + (1/7*1/6) + (1/7*1/5)+(1/6*1/5)

?

crap

change it to


(((1 / 16) * 1) / 15) + (((1 / 16) * 1) / 14) + (((1 / 16) * 1) / 13) + (((1 / 15) * 1) / 14) + (((1 / 15) * 1) / 13) + (((1 / 14) * 1) / 13) = 0.0288232601

[Jay Hova]

actually forget it:|

[CSR]

good try guys

[kingsley]

Use combinatorics.

hahah that was funny. I did some lulz.

[Jay Hova]

crap

change it to


(((1 / 16) * 1) / 15) + (((1 / 16) * 1) / 14) + (((1 / 16) * 1) / 13) + (((1 / 15) * 1) / 14) + (((1 / 15) * 1) / 13) + (((1 / 14) * 1) / 13) = 0.0288232601

actually, i think this should be right.

picking 4 shoes, we can either have 1-2, 1-3, 1-4, 2-3, 2-4, 3-4.
1-2: 1st shoe is "right" AND (hence the *) 2nd shoe (1/15) is "right"
OR (+)
1-3: 1st shoe is "right" AND (*) the 3rd shoe (1/14)
so on and so forth.

[Jay Hova]

And if you really want to know....

you could write a VB or any other app..that could solve this.

if you run it for 10,000 to 100,000 times...it'll eventually converge to it's theoretical probability. :lol::cry:

[EmperorOfCanada]

actually, i think this should be right.

picking 4 shoes, we can either have 1-2, 1-3, 1-4, 2-3, 2-4, 3-4.
1-2: 1st shoe is "right" AND (hence the *) 2nd shoe (1/15) is "right"
OR (+)
1-3: 1st shoe is "right" AND (*) the 3rd shoe (1/14)
so on and so forth.

I really honestly have no idea what you are talking about... Explain why yours is right and mine is wrong?

[jzy]

Q1: ((1/16*1/15)*((1/14*12/13)*2*7))*8=0.030769

Q2:

15c5=3003

6c3=20

9c2=36

(20*36)/3003=0.23976

[Jay Hova]

I really honestly have no idea what you are talking about... Explain why yours is right and mine is wrong?

If you pick out 4 shoes. to create a match you can have the following cases

1st shoe and 2nd shoe match
1st shoe and 3rd shoe match
1,4
2,3
3,4

To get the first situation: The probability of us getting the right shoe is 1/16, AND, to get the next exact shoe, is 1/15.
(+) The next situation: it's 1/16 AND 1/14 (cause the shoe prior wasn't it).

[Jon Lai]

hahah that was funny. I did some lulz.

Err.. why?

This is obviously a question from Gr12 Data Management Unit 4: Probability, which introduces the use of permutations and combinatorics, which is exactly how you're supposed to solve the question.