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Freaking Statistics...
Hey guys. I've never been good with statistics... I can easily handle calculus, integration, differentials, geometry, trig and algebra, but apparently the subtleties of statistics are too minute for me to understand. I had some experience with statistics in my calculus 2 class, and that is the only test I've ever failed in any math class. I'm taking a course in genetics, and obviously statistics are an important part of the class. I need help with the following questions, and I would appreciate any suggestions as well to help me overcome this problem. I understand the basics, but apparently that's not enough...
1. What is the probability of throwing two dice and obtaining a 4 and a 6? I'm getting 1/36, because the P of getting any side is 1/6, and because the conditions are independent, I multiplied them...But the correct answer is 1/18. My mind is boggling... 2. A family has five kids. What is the probability that the first and last born are male? I have no clue how to get this. Obviously, the probability of getting a single male or female is 1/2, but I can't handle the order... What do I do? The answer is 1/4. 3. What is the probability of flipping a coin six times in a row and getting three heads in a row, followed by three tails? What do I do with this? The answer's 1/64. 4. What is the probability of flipping a coin six times in a row and getting three heads in a row and three tails? Once again, how do I account for the order? The answer's 5/16. Any help would be greatly appreciated. |
I'm in a stats class right now, so helping you with this helps me prepare for my midterm on Monday night! :) You can PM me if these answers don't make sense to you.
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Oh, I get No.2 now... I thought it also wanted to take into account the 3 others in between the first and last, but now it makes sense. Thanks for No.3, that makes sense. And I think for No.4, it's asking what the probability of getting 3 heads in a row is... You can either get 3 h's then 3 t's, or 1 t then 3 h's then 2 t's, 2 t's then 3 h's then 1 t, or 3 t's then 3 h's. I know I'd have to find the P of each case then add them up, but I don't know how to take the orders into account.
Thanks a lot. :) |
On #1, you're both forgetting that it doesn't matter which die has which number. So it's 1/6 x 1/6 x 2. If the numbers were die-specific, you'd have the right answer.
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Thanks, Jazz. That's really helpful.
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Hmm... But the P of getting two 3's is 1/36...How's that different than #1? If the P of getting a 3 on one die is 1/6, then so is the P of getting a 4 on another...So why multiply by 2 when you're seeking the P for two different numbers, and not when seeking the P for the same number?
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Because of the 36 ways the two dice could fall, TWO satisfy the condition that you get a four and a six: die one has a four and die two has a six OR die one has a six and die two has a four. That's two chances out of 36, which simplifies to 1/18.
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You could do this longhand too, if you really wanted to get a tactile feel for this sort of problem. You could make a chart figuring out all of the possible orders in which six coin flips would come out. You'd find that there are 64 possible ways those six flips could turn out, of which only ONE would be "HHHTTT". Quote:
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See I like probabilities like this, it's like gambling. Real statistics with poisson distributions and deviations and all that other flying mess makes my head spin. Too much memorization with equations and tables that were entirely hand-waved into existence. |
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I agree; I believe as stated, the answer is 1/16. |
OK I just asked the world's greatest math geek and he agrees that the answer to #4 is 1/16, making the answer on the paper wrong.
EDIT: I would also like to add something about #3. Don't be fooled by a question like this: If you flipped a coin ten times, which H/T pattern is more likely:They are both equally likely because it is a specific pattern. Just because B looks more random does not make it specifically more likely. Getting one that APPEARS TO BE RANDOM is more likely however. |
Well how did you do on the assignment?
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