Bored? Come in and play at Jusunlee.com Arcade! Go chat in Jusunlee.com Chatroom (requires AIM) Here you can view your subscribed threads, work with private messages and edit your profile and preferences Registration is free! Calendar Find other members Frequently Asked Questions Search Home
Jusunlee.com Forums > Education > Homework Help > If you're good at adv. alg..
  Last Thread   Next Thread
Author
Thread Post New Thread    Post A Reply
TAIgrr
MM caramel FRAP! =P

Registered: Mar 2002
Location:
Posts: 5644
Status: Offline

If you're good at adv. alg..

PLEASE HELP ME !!

ON PERMUATIONS, PROBABILITIES, COMBINATIONS..

I NEED THE SET-UP.. NOT JUST THE ANSWER

1) a coin purse contains 5 pennies, 6 nickels, 7 dimes. 3 coins are selected at random without replacement. find the probability that all 3 coins are dimes.

2) from a standard deck of 52 cards, 5 cards are dealt. what are the odds that all 5 cards are red?

3) from a group of 8 men and 10 women, a committee of 3 is to be selected at random. find the probability that all 3 are men or all 3 are women

4) 4 fair coins are tossed. find the probability that at least 3 show heads.

5) determine all real values of w for which the equation C(w+1, 2) = 9C(w, 1) is true.

__________________
.

Report this post to a moderator | IP: Logged

Old Post 04-11-2003 05:28 AM
Click Here to See the Profile for TAIgrr Click here to Send TAIgrr a Private Message Find more posts by TAIgrr Add TAIgrr to your buddy list Edit/Delete Message Reply w/Quote
Chinesegrl
Senior Member

Registered: Aug 2002
Location: Nevada
Posts: 1444
Status: Offline

Re: If you're good at adv. alg..

quote:
Originally posted by TAIgrr
[B]PLEASE HELP ME !!

ON PERMUATIONS, PROBABILITIES, COMBINATIONS..

I NEED THE SET-UP.. NOT JUST THE ANSWER

1) a coin purse contains 5 pennies, 6 nickels, 7 dimes. 3 coins are selected at random without replacement. find the probability that all 3 coins are dimes.



you jus add them so it equals.. 18 and you want 3 so.. 3/18 its 6% i think >.< hm.. or you find the percentage of 3 from 7 and then you find the percentage of 7 out of 18 then i think you either subtract or add.. i think subtract well im only using my best guessin..

Report this post to a moderator | IP: Logged

Old Post 04-11-2003 05:49 AM
Click Here to See the Profile for Chinesegrl Click here to Send Chinesegrl a Private Message Visit Chinesegrl's homepage! Find more posts by Chinesegrl Add Chinesegrl to your buddy list Edit/Delete Message Reply w/Quote
s0lotu
Member

Registered: Mar 2003
Location:
Posts: 63
Status: Offline

#1

no, no. let me explain:

total = 18. then when you first pick the chance of getting a dime is 3/18. so then now there are 17 left since it's without replacement. now your chance of getting a dime is 2/17. for the same reason, for the third pick, your chance is 1/16. (3/18)(2/17)(1/16) = 6/4896

Report this post to a moderator | IP: Logged

Old Post 04-11-2003 06:02 AM
Click Here to See the Profile for s0lotu Click here to Send s0lotu a Private Message Find more posts by s0lotu Add s0lotu to your buddy list Edit/Delete Message Reply w/Quote
s0lotu
Member

Registered: Mar 2003
Location:
Posts: 63
Status: Offline

#2

same deal here.
total = 52 cards, of which 26 are red when you first start.
so for the first pick, the chance is 26/52 (or 1/2)
now ask yourself, what is the probability of picking a red on the next one? (how many are left in total, and how many reds are left?)
you try this one on ur own

Report this post to a moderator | IP: Logged

Old Post 04-11-2003 06:08 AM
Click Here to See the Profile for s0lotu Click here to Send s0lotu a Private Message Find more posts by s0lotu Add s0lotu to your buddy list Edit/Delete Message Reply w/Quote
s0lotu
Member

Registered: Mar 2003
Location:
Posts: 63
Status: Offline

#3

ok, same here, except there's the OR involved.
first find the probability that all 3 are men. then the problem is like #1 or #2. total of 18 people, 8 of which are men... etc.
in the same way, then find the probability that all 3 are women.
then since it says OR, you add them. (explanation: if you have 5 shirts of all different colors and you wanna wear a shirt that is pink OR green, when you randomly grab a shirt from the closet without looking, you have 2 chances out of 5 that you will grab one that u wanna wear (pink OR green)).

Report this post to a moderator | IP: Logged

Old Post 04-11-2003 06:19 AM
Click Here to See the Profile for s0lotu Click here to Send s0lotu a Private Message Find more posts by s0lotu Add s0lotu to your buddy list Edit/Delete Message Reply w/Quote
TAIgrr
MM caramel FRAP! =P

Registered: Mar 2002
Location:
Posts: 5644
Status: Offline

Re: #1

quote:
Originally posted by s0lotu
no, no. let me explain:

total = 18. then when you first pick the chance of getting a dime is 3/18. so then now there are 17 left since it's without replacement. now your chance of getting a dime is 2/17. for the same reason, for the third pick, your chance is 1/16. (3/18)(2/17)(1/16) = 6/4896



but the answer is suppose to come out to be 35/316...

__________________
.

Report this post to a moderator | IP: Logged

Old Post 04-11-2003 06:19 AM
Click Here to See the Profile for TAIgrr Click here to Send TAIgrr a Private Message Find more posts by TAIgrr Add TAIgrr to your buddy list Edit/Delete Message Reply w/Quote
s0lotu
Member

Registered: Mar 2003
Location:
Posts: 63
Status: Offline

oh oops.
there are 7 dimes to start
so (7/18)(6/17)(5/16) = 35/816

Report this post to a moderator | IP: Logged

Old Post 04-11-2003 06:23 AM
Click Here to See the Profile for s0lotu Click here to Send s0lotu a Private Message Find more posts by s0lotu Add s0lotu to your buddy list Edit/Delete Message Reply w/Quote
TAIgrr
MM caramel FRAP! =P

Registered: Mar 2002
Location:
Posts: 5644
Status: Offline

quote:
Originally posted by s0lotu
oh oops.
there are 7 dimes to start
so (7/18)(6/17)(5/16) = 35/816


... but the answer IS.. 316 not 816

(i have all the answers, but i just need to know how to GET them..)

__________________
.

Report this post to a moderator | IP: Logged

Old Post 04-11-2003 06:24 AM
Click Here to See the Profile for TAIgrr Click here to Send TAIgrr a Private Message Find more posts by TAIgrr Add TAIgrr to your buddy list Edit/Delete Message Reply w/Quote
s0lotu
Member

Registered: Mar 2003
Location:
Posts: 63
Status: Offline

that's wrong. it's 816.

Report this post to a moderator | IP: Logged

Old Post 04-11-2003 06:25 AM
Click Here to See the Profile for s0lotu Click here to Send s0lotu a Private Message Find more posts by s0lotu Add s0lotu to your buddy list Edit/Delete Message Reply w/Quote
s0lotu
Member

Registered: Mar 2003
Location:
Posts: 63
Status: Offline

here, check the next one:
if you do the same thing for #2, i get 253/9996

Report this post to a moderator | IP: Logged

Old Post 04-11-2003 06:33 AM
Click Here to See the Profile for s0lotu Click here to Send s0lotu a Private Message Find more posts by s0lotu Add s0lotu to your buddy list Edit/Delete Message Reply w/Quote
s0lotu
Member

Registered: Mar 2003
Location:
Posts: 63
Status: Offline

#4

anyway, let me go on:
so the next one, u want at least 3 to show heads.
so you basically want to know what the probability that 3 out of 4 will show heads OR all will be heads.

for all heads, it's easy. there is only one way to get all heads: TTTT. each coin that is tossed has a 1/2 probability of being heads. since there are 4 coins, (1/2)(1/2)(1/2)(1/2) = 1/16

for 3 heads, you have to note that there are 4 ways of getting this:
1) HTTT
2) THTT
3) TTHT
4) TTTH
this one is easy to see, but this is when you would use combinations. if you do the combination C(4,3) since u want 3 heads out of 4 you will get 4. note that C(4,3) is the same as C(4,1) since it is the same as the combinations when there is 1 T out of 4, as u can see in the list above. in notation, C(n,k) = C(n,n-k)

anyway, that's all the combinations of getting 3 heads. now you have to compute the probability. for each combination, it's the same. since the probability of heads = probability of tails = 1/2, it's (1/2)(1/2)(1/2)(1/2) again, or 1/16. then remember, there were 4 different ways, so 4/16.

so the probability of at least 3 heads = probability of all heads + probability of 3 heads = 1/16 + 4/16 = 5/16

Report this post to a moderator | IP: Logged

Old Post 04-11-2003 06:47 AM
Click Here to See the Profile for s0lotu Click here to Send s0lotu a Private Message Find more posts by s0lotu Add s0lotu to your buddy list Edit/Delete Message Reply w/Quote
s0lotu
Member

Registered: Mar 2003
Location:
Posts: 63
Status: Offline

oops
in the last one, i switched heads and tails so i meant
1) HHHT
2) HHTH
3) HTHH
4) THHH

Report this post to a moderator | IP: Logged

Old Post 04-11-2003 06:48 AM
Click Here to See the Profile for s0lotu Click here to Send s0lotu a Private Message Find more posts by s0lotu Add s0lotu to your buddy list Edit/Delete Message Reply w/Quote
s0lotu
Member

Registered: Mar 2003
Location:
Posts: 63
Status: Offline

haha and i meant HHHH for all heads. sorry, it's late.

Report this post to a moderator | IP: Logged

Old Post 04-11-2003 06:48 AM
Click Here to See the Profile for s0lotu Click here to Send s0lotu a Private Message Find more posts by s0lotu Add s0lotu to your buddy list Edit/Delete Message Reply w/Quote
s0lotu
Member

Registered: Mar 2003
Location:
Posts: 63
Status: Offline

5

aiite, last one

C(n,k) = n!/(k!(n-k)!)

so
C(w+1, 2) = (w+1)!/(2!(w-1)!)
9C(w,1)=9w!/(1!(w-1)!)

set them equal to each other. multiply both sides by (w-1)! and they cancel out from each equation. now you have:

(w+1)!/2 = 9w!

since (w+1)! = (w+1)(w)(w-1)(w-2)...(1)
and w! = (w)(w-1)(w-2)...(1)
divide both sides by w!, and everything starting from w..1 cancels out on both sides

you're left with
(w+1)/2=9
so w+1=18 and w=17

Report this post to a moderator | IP: Logged

Old Post 04-11-2003 07:01 AM
Click Here to See the Profile for s0lotu Click here to Send s0lotu a Private Message Find more posts by s0lotu Add s0lotu to your buddy list Edit/Delete Message Reply w/Quote
s0lotu
Member

Registered: Mar 2003
Location:
Posts: 63
Status: Offline

k, good luck on ur hw
i have a lab due tomorrow, gonna be up all night
hehe, yea, i'm procrastinating rite now
you can PM me if u have any questions about what i did

Report this post to a moderator | IP: Logged

Old Post 04-11-2003 07:04 AM
Click Here to See the Profile for s0lotu Click here to Send s0lotu a Private Message Find more posts by s0lotu Add s0lotu to your buddy list Edit/Delete Message Reply w/Quote
TAIgrr
MM caramel FRAP! =P

Registered: Mar 2002
Location:
Posts: 5644
Status: Offline

quote:
Originally posted by s0lotu
k, good luck on ur hw
i have a lab due tomorrow, gonna be up all night
hehe, yea, i'm procrastinating rite now
you can PM me if u have any questions about what i did


haha.. thanks thanks.. but (i got off the comp before you started posting the last 5 replies. -_-

haha, thanks tho.

__________________
.

Report this post to a moderator | IP: Logged

Old Post 04-11-2003 11:37 PM
Click Here to See the Profile for TAIgrr Click here to Send TAIgrr a Private Message Find more posts by TAIgrr Add TAIgrr to your buddy list Edit/Delete Message Reply w/Quote
MaGiKToToRo
~dance~

Registered: Jan 2003
Location: New York
Posts: 740
Status: Offline

ahhh.... probability

__________________
"Fear less, hope more;
Whine less, breathe more;
Talk less, say more;
Hate less, love more;
And all good things are yours."


-Sweedish Proverb


aim: MaGiKToToRo
xanga: MaGiKToToRo

Report this post to a moderator | IP: Logged

Old Post 05-07-2003 03:41 AM
Click Here to See the Profile for MaGiKToToRo Click here to Send MaGiKToToRo a Private Message Visit MaGiKToToRo's homepage! Find more posts by MaGiKToToRo Add MaGiKToToRo to your buddy list Edit/Delete Message Reply w/Quote
All times are GMT. The time now is 12:54 AM. Post New Thread    Post A Reply
  Last Thread   Next Thread

Show Printable Version Email this Page Subscribe to this Thread

Forum Rules:
You may not post new threads
You may not post replies
You may not post attachments
You may not edit your posts
HTML code is OFF
vB code is ON
Smilies are ON
[IMG] code is ON