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TAIgrr
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 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. __________________ TAIgrr's Xanga ! Report this post to a moderator | IP: Logged
04-11-2003 05:28 AM
Chinesegrl
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 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
04-11-2003 05:49 AM
s0lotu
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 #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
04-11-2003 06:02 AM
s0lotu
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 #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
04-11-2003 06:08 AM
s0lotu
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 #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
04-11-2003 06:19 AM
TAIgrr
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 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... __________________ TAIgrr's Xanga ! Report this post to a moderator | IP: Logged
04-11-2003 06:19 AM
s0lotu
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 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
04-11-2003 06:23 AM
TAIgrr
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 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..) __________________ TAIgrr's Xanga ! Report this post to a moderator | IP: Logged
04-11-2003 06:24 AM
s0lotu
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 that's wrong. it's 816. Report this post to a moderator | IP: Logged
04-11-2003 06:25 AM
s0lotu
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 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
04-11-2003 06:33 AM
s0lotu
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04-11-2003 06:47 AM
s0lotu
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 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
04-11-2003 06:48 AM
s0lotu
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 haha and i meant HHHH for all heads. sorry, it's late. Report this post to a moderator | IP: Logged
04-11-2003 06:48 AM
s0lotu
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 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
04-11-2003 07:01 AM
s0lotu
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 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
04-11-2003 07:04 AM
TAIgrr
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 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. __________________ TAIgrr's Xanga ! Report this post to a moderator | IP: Logged
04-11-2003 11:37 PM
MaGiKToToRo
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 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
05-07-2003 03:41 AM
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