JSngry Posted September 3, 2004 Report Posted September 3, 2004 (edited) Freshman Algebra (high school) What is the probability of drawing 3 spades out of a deck of cards without replacement? My daughter says the answer = (13/52)*(12/51)*(11/50)*(10/49) She gets an answer of some ungodly fraction. None of this "sits right" with what I remember of my various math classes. I say there's got to be an easier way, but darned if I can remember what it is. This ain't the kind of thing I do on a daily basis, doncha' know... As always, and especially this time, thanks in advance! Edited September 3, 2004 by JSngry Quote
ejp626 Posted September 3, 2004 Report Posted September 3, 2004 Freshman Algebra (high school) What is the probability of drawing 3 spades out of a deck of cards without repalcement? My daughter says the answer = (13/52)*(12/51)*(11/50)*(10/49) She gets an answer of some ungodly fraction. None of this "sits right" with what I remember of my various math classes. I say there's got to be an easier way, but darned if I can remember what it is. This ain't the kind of thing I do on a daily basis, doncha' know... As always, and especially this time, thanks in advance! This is close, but she needs three terms not four, so she would stop at 11/50, which will still be a very small fraction. Here she is calculating drawing four spades. Quote
JSngry Posted September 3, 2004 Author Report Posted September 3, 2004 Thanks! Those fractions reduce down real nicely, too. Nice, easy answer. Probablility formulas went south for me a LONG time ago... Quote
bertrand Posted September 3, 2004 Report Posted September 3, 2004 You can derive the answer using the formula for conditional probability: P (A given B) = P (A and B)/P (B). If this still doesn't help, I'll explain more later. Gotta go read the cat in the hat to someone now. Bertrand. Quote
Big Al Posted September 3, 2004 Report Posted September 3, 2004 Isn't it like this: on the first draw, she has a 1:13 chance; on the second draw, a 1:17 chance; and on the third draw, a 1:25 chance. So unless I'm thinking this incorrectly, she has a 1:(13*17*25) chance, or 1:5525 chance of drawing three consecutive spades. I'm assuming it's consecutive. Is that part correct? Or does she put the card back in the deck after drawing it? Quote
king ubu Posted September 3, 2004 Report Posted September 3, 2004 Now this is one of the most amusing threads I've seen on this board! Can't offer any help, as these dreaded formulas never stuck to my head (had a programmable calculator for such stuf...) Quote
JohnJ Posted September 3, 2004 Report Posted September 3, 2004 is the deck complete? Also, what about the Joker? Quote
Guy Berger Posted September 3, 2004 Report Posted September 3, 2004 Freshman Algebra (high school) What is the probability of drawing 3 spades out of a deck of cards without replacement? My daughter says the answer = (13/52)*(12/51)*(11/50)*(10/49) She gets an answer of some ungodly fraction. None of this "sits right" with what I remember of my various math classes. I say there's got to be an easier way, but darned if I can remember what it is. This ain't the kind of thing I do on a daily basis, doncha' know... As always, and especially this time, thanks in advance! 100%, if you keep drawing until you run out of cards... Guy Quote
Shrdlu Posted September 3, 2004 Report Posted September 3, 2004 Ah, the old deck of cards problems in probability! There would be little in those courses if you took those out. I wish I had a penny for the number of times I have taught that! A friend of mine had a great pun on this stuff: "One man's mean is another man's Poisson!" By the way, I agree with ejp626's answer. Quote
Man with the Golden Arm Posted September 3, 2004 Report Posted September 3, 2004 last time i tried this i got taken for a five by some cats w/ a cardboard box. and here i was fretting about upcoming fourth grade rithmetic. Quote
Man with the Golden Arm Posted September 3, 2004 Report Posted September 3, 2004 But really I think you got something here. How about a dedicated forum to figuring out all things homework related? ...and when the kiddies come to school they can say that jazz changes all things for the positive!!! Quote
JSngry Posted September 3, 2004 Author Report Posted September 3, 2004 Isn't it like this: on the first draw, she has a 1:13 chance; on the second draw, a 1:17 chance; and on the third draw, a 1:25 chance. So unless I'm thinking this incorrectly, she has a 1:(13*17*25) chance, or 1:5525 chance of drawing three consecutive spades. Al, how do you get that 1:25? Quote
JSngry Posted September 3, 2004 Author Report Posted September 3, 2004 is the deck complete? Many is the time that either I or those I work with have been accused of playing with less than a full deck.... Quote
JSngry Posted September 3, 2004 Author Report Posted September 3, 2004 is the deck complete? Also, what about the Joker? See above. Quote
JSngry Posted September 3, 2004 Author Report Posted September 3, 2004 (edited) How about a dedicated forum to figuring out all things homework related? Don't know about an entire forum, but tell you what - we got so many edjumacated individuals from so many different fields here that considering the Organissimo Board as a potential "homework help resource" for flusterated parents is NOT a bad idea. Especially since it seems ike there's somebody here 24-7! Edited September 3, 2004 by JSngry Quote
Big Al Posted September 3, 2004 Report Posted September 3, 2004 Isn't it like this: on the first draw, she has a 1:13 chance; on the second draw, a 1:17 chance; and on the third draw, a 1:25 chance. So unless I'm thinking this incorrectly, she has a 1:(13*17*25) chance, or 1:5525 chance of drawing three consecutive spades. Al, how do you get that 1:25? Sheer stupidity, pure & simple. Y’know how when you’re about to fall asleep, you have some kind of mind-boggling revelation, but you’re either too tired or too lazy to get up and do anything about it? Well, right before I fell asleep last night, it occurred to me that I answered that problem with the assumption that she was drawing three ACES (of which there are only four), as opposed to three SPADES (of which there are thirteen). Unfortunately, when I realized this, I was almost asleep and the computer was shut down. So, had the problem been for three ACES, then she would’ve had a four-in-52 chance (1:13) on her first draw; a three-in-51 (1:17) on her second draw; and a two-in-50 (1:25) on her third draw. Scary thing is, I’m an accountant by day. Quote
frank m Posted September 3, 2004 Report Posted September 3, 2004 Stop using formulas and just use logic. The previous post is almost right, but wrong. First chance is 1/13 ok. second is 3/51. third chance is 2/50----but since it doesn't matter what order you draw them the result is too big by a factor of the number of ways you can rearrange the 3 picked cards and still have the same group of three. So the correct answer is 1/13* 3/51* 2/5 _____________ 3*2*1 Quote
Shrdlu Posted September 3, 2004 Report Posted September 3, 2004 In a situation with a finite number of outcomes, you calculate the probability of a particular kind of outcome by the general formula P = (number of outcomes of that particular kind)/(total number of outcomes), where the / means "divided by". (Sorry, I have no access to mathematical symbols here.) In the case of the three spades, drawn randomly without replacement, there are 13 x 12 x 11 outcomes with three spades, and 52 x 51 x 50 outcomes in all (that is, any three cards). So, we have P = (13 x 12 x 11)/(52 x 51 x 50), as was said above. If you have any questions about this, grab any book on Introductory Probability and Statistics. A lot of the freshman "Calculus and Analytic Geometry" books have a chapter on this, too. Quote
BruceH Posted September 3, 2004 Report Posted September 3, 2004 Is she drawing from the bottom of the deck? Quote
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