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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.

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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..

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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

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Old Post 04-11-2003 06:02 AM
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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

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Old Post 04-11-2003 06:08 AM
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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)).

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Old Post 04-11-2003 06:19 AM
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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...

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Old Post 04-11-2003 06:19 AM
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s0lotu
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oh oops.
there are 7 dimes to start
so (7/18)(6/17)(5/16) = 35/816

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Old Post 04-11-2003 06:23 AM
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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..)

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Old Post 04-11-2003 06:24 AM
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s0lotu
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that's wrong. it's 816.

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Old Post 04-11-2003 06:25 AM
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s0lotu
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here, check the next one:
if you do the same thing for #2, i get 253/9996

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Old Post 04-11-2003 06:33 AM
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s0lotu
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#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

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Old Post 04-11-2003 06:47 AM
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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

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Old Post 04-11-2003 06:48 AM
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s0lotu
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haha and i meant HHHH for all heads. sorry, it's late.

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Old Post 04-11-2003 06:48 AM
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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

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Old Post 04-11-2003 07:01 AM
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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

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Old Post 04-11-2003 07:04 AM
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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.

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Old Post 04-11-2003 11:37 PM
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MaGiKToToRo
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ahhh.... probability

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