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By M660 on Saturday, December 28, 2002 - 10:30 pm: Edit |

A)How do you find the sum of all integers between 80 and 110?

B)first the ratio of males to females at F. College was 2:3.Since then enrollment has gone up by 400 and the ratio of female students is the same.The ratio of males to females is 1:1. HOW many students are currently enrolled at F. College??

VERY IMPORTANT!..PS..i think probability is the devil....

c.)in a stack of six cards, each card is labeled with a different integer 0 through 5. If two cards are selected at random w/o replacement, what is the probability that their sum will be two?

By 800m on Saturday, December 28, 2002 - 10:40 pm: Edit |

A) sum=(80+110)*31/2=2945 (there are 31 numbers from 80 to 110)

B)number of current students: x

number of current males: 2x/5

females: 3x/5

after the increase in enrollment:

(400+x)/2=3x/5

x=2000

C)there are 6*5=30 ways to choose the cards.

three ways to get a sum of two {0,2},{2,0},and {1,1}

so the probability is 3/30 or 1/10

P.S. I'm really TERRIBLE at prob, so don't trust me on that.

By hey hey hey on Saturday, December 28, 2002 - 10:57 pm: Edit |

you are wrong on #2 and #3, the answer for #2 is 2400 and the answer for #3 is 2/15

By SHAME on Saturday, December 28, 2002 - 11:04 pm: Edit |

sorry, by "current student", I meant the enrollement before the increase. so i should have added 400 to 2000 to make it 2400. didn't read that carefully.

i guess for #3 i shouldn't have counted {0,2} and {2,0} twice. so that would be 2/15.

hmmm, getting so many problems wrong i don't think i deserve an 800...if today were the test day, i would have been like dead

By SHAMEnever answering again on Saturday, December 28, 2002 - 11:06 pm: Edit |

no for #3 i actually missed a {1,1} damn it!!!

By Hello on Saturday, December 28, 2002 - 11:07 pm: Edit |

I think that you just forgot to add a 400 to the answer...

By Pat57575 (Pat57575) on Saturday, December 28, 2002 - 11:49 pm: Edit |

lol, 800m, you can't pull two 1's... each of the cards is numbered differently and the first draw isn't replaced.

You could either pull a "0" (1/6 chance) first and then a "2" (1/5 chance). The odds of that happening are (1/6)(1/5)=1/30

Or you could pull a "2" (1/6 chance) first and then a "0" (1/5 chance). The odds of that happening are also 1/30

The odds of either scenario happening are (1/30)+(1/30)= 1/15

By SHAME on Sunday, December 29, 2002 - 12:09 am: Edit |

but how come the answer is 2/15?

By M660 on Sunday, December 29, 2002 - 12:46 am: Edit |

Thanks again..

as you can see, im no fan of probability. Is there a good source online that has good tips or is there some fundamental rule that im unaware of?

my goal is to break 700 on the math...hopefully reach 750. I only get the hard ones wrong, and maybe an occasional medium.

By Pat57575 (Pat57575) on Sunday, December 29, 2002 - 03:20 pm: Edit |

i dunno, maybe the book is wrong... I'm pretty confident about that answer

By The_Sagacious_Owl on Sunday, December 29, 2002 - 08:59 pm: Edit |

M660, when you post on the internet and inquire about questions; please provide the correct information. The sum of the numbers is 3, this caused confusion among the posters.

The correct question should read as so. In a stack of six cards, each card is labled with a different integer 0 through 5. If two cards are selected at random without replacement, what is the probability that their sum will be 3.

There are 30 possible outcomes, however only 2 will yield an answer of 3. These are 3,0 and 2,1, thereby making the answer 2/15. If the sum of the numbers was 2, then indeed the probability would be 1/30 as was indicated by Pat.

Hope this helped.

By M660 on Sunday, December 29, 2002 - 09:28 pm: Edit |

I KNOW I SCREWED UP! under "probability again" i apologized. sorry.. thanks tho

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