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Nytefish
United Kingdom4282 Posts
nCr means the number of ways of choosing r objects from n objects, where order doesn't matter. How does the number of ways of choosing 2 cases from 8 help? (it doesn't) I gave you the answer in the last line of my previous post O_o. And in the one before that told you that nCr and nPr aren't needed. @monoxide I know you got the answer but he clearly doesn't understand why  | ||
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OmgIRok
Taiwan2699 Posts
OH wAIT YEAAA Okay so the success rates are all the positions where its 1XXXXXXX8 and since 1 & 8 are fixed you just find how many possible ways you can arrange 23456 which is 6! YES I SEE WHAT YOU DID THERE | ||
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OmgIRok
Taiwan2699 Posts
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Nytefish
United Kingdom4282 Posts
The later ones will need you to use nCr and nPr? | ||
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OmgIRok
Taiwan2699 Posts
On February 23 2009 10:26 Nytefish wrote: Yeah, these are the "easy" questions right? The later ones will need you to use nCr and nPr? Um... I think we are supposed to use ncr and npr on all of them as a shortcut to the counting principle but yeah whatever cuz i know npr and ncr formulas [not exactly sure what they are used for yet though, pretty sure used for what i mentioned earlier] | ||
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OmgIRok
Taiwan2699 Posts
I already did some of them, please check my answers? XD 3 people are chosen randomly out of 3 sophomores and 3 juniors #1 P(0 sophmores) 3C0 / 6C3 = 1/20 2 P(1 sophmore) 3C1 / 6C3 = 3/20 #3 P(2 sophomore) 3 C 2 / 6C3 = 3/20 #4 P(3 sophomores) 3C3 / 6C3 = 1/20 #5 P(2 juniors) Currently on #5 | ||
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huameng
United States1133 Posts
They don't add up because there are more than 3 ways to make a group with 1 or 2 sophomores. You also have to account for the number of different ways to pick the juniors. | ||
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OmgIRok
Taiwan2699 Posts
okay I found out that P(1) and p(2) sophomores is actually supposed to be 9/20 chance to be picked, but where is that logic found? 6C1 gives me 6 6C2 gives me 15 15-6 = 9  | ||
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Malongo
Chile3472 Posts
#5 is easy if you use the same its P(2)=3C1*3C2/6C3=9 | ||
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OmgIRok
Taiwan2699 Posts
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Malongo
Chile3472 Posts
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OmgIRok
Taiwan2699 Posts
On February 23 2009 11:15 malongo wrote: I note a lack of understanding in the multiplication principle when counting. The basic is that if an event A can occur in a different ways and an event B can occur in b different ways THEN the numbers of ways that the event A and B can occur is a*b. haha yeah totally forgot about counting principle X_X eff me | ||
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Abydos1
United States832 Posts
On February 23 2009 09:00 OmgIRok wrote: #1. P(Seasons 1-4 in the correct positions) So 1234XXXX I got 1/70 for this one. I seriously doubt it's correct though. #2. P(Seasons 1 and 8 in the correct positions) 1XXXXXX8 Currently stuck on this one. Does it specifically state those are the only ones in the correct positions? Because P(1 & 8 correct) can have other seasons in correct spots the way you worded it. | ||
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DeathByMonkeys
United States742 Posts
8 nPr 5 or 8x7x6x5 the second one is 8 nPr 1 or 8! or 8x7x6x5x4x3x2x1 Each problem follows the same principles; you have a 1/8 chance of randomly placing the first dvd correctly, then a 1/7 chance of the next dvd and so on... | ||
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Malongo
Chile3472 Posts
The birthday problem: In a room there are n people, what is the probability that at least 2 of them have a common birthday? hint + Show Spoiler + use the complement probability, that is find the P(no pair of them have the same birthday), then P(at least 2 of them have a common birthday)= 1-P(no pair of them have the same birthday) solution + Show Spoiler + since there are 365 posible birthdays then the total number of cases is 365!. The number of ways that no posible repetition of birthdays occur is 365*364*...*(365-n+1) because the first one have 365 posible birthdays, the second has 364 posible birthdays (because we want that he doesnt have the same birthday as the first), the third 363 and so on. then P(no common birthday)= 365*364*...*(365-n+1)/365!=(365Cn)*n!/365! and the desired probablity is 1-(365Cn)*n!/365!. why is this problem interesting? because its highly counterintuitive, for n=40 P>0.5 so for n=90 its almost sure that there are 2 with the same birthday. | ||
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JeeJee
Canada5652 Posts
check your equations i think you typo'd something and 23 is the magic number not 40 last i checked | ||
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Monoxide
Canada1190 Posts
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Malongo
Chile3472 Posts
On February 23 2009 13:33 JeeJee wrote: @malongo check your equations i think you typo'd something and 23 is the magic number not 40 last i checked ?? please what is wrong? i rechecked but cant find it. And for the actual numebers i just put numbers that i was sure to fit, i dont remember them from memory i took probabilities about3 years ago dont ask that much ahaha | ||
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