archetypal fool

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.

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PonyPotato

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.

Honestly, I would answer 1/6 * 1/6 = 1/36 on this question as well. There may just be a difference because you're rolling two dice at the same time. Hmm. I might crack my stats book on this one, I need to study anyway.

The order has no bearing on the answer of the question; the events are independent and the probability of 1/2 will not change based on the sex of the other children. You just want the probability of getting two males, which is 1/2 * 1/2 = 1/4.

The probability of getting heads (or tails) is 1/2. You just multiply it out for six coin flips, 1/2 * 1/2 * 1/2 * 1/2 * 1/2 * 1/2 = 1/64 (or just do [1/2]^6). It doesn't matter whether you're getting heads or tails on each flip, the events are independent.

This is the exact same as question 3? Can you check the question and get back to me?

Do I get a cookie? I hope I helped at least a little bit.

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archetypal fool

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.

__________________
I have my own particular sorrows, loves, delights; and you have yours. But sorrow, gladness, yearning, hope, love, belong to all of us, in all times and in all places. Music is the only means whereby we feel these emotions in their universality. ~H.A. Overstreet

The_Jazz

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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PonyPotato

Thanks, Jazz. That's really helpful.

__________________
"Without passion man is a mere latent force and possibility, like the flint which awaits the shock of the iron before it can give forth its spark."
— Henri-Frédéric Amiel

archetypal fool

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?

__________________
I have my own particular sorrows, loves, delights; and you have yours. But sorrow, gladness, yearning, hope, love, belong to all of us, in all times and in all places. Music is the only means whereby we feel these emotions in their universality. ~H.A. Overstreet

ratbastid

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.

The kids in the middle are a red herring. The question really is: given two kids, what are the odds they're both male. Probabilities multiply when you're talking about a string of outcomes like this, so we're talking 1/2 * 1/2 = 1/4.


Again, a string of outcomes means you multiply the probabilities of each individual (in this case) flip. (1/2)^6 = 1/64.

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".

Not the way you've stated it here, it isn't. The way you've stated it, it's the same problem as #3, and the answer is 1/64. And I hope my answer on #3 served to back you down off the ledge about "accounting for the order". There's no need--multiply the odds of each individual element of the series, and you've got your odds for the series.

Hain

1. What is the probability of throwing two dice and obtaining a 4 and a 6?

   Think of it like this- when you roll, you roll them one at a time. The first die can be either a 4 or a 6, so two outcomes out of six are preferred- 2/6 or 1/3. The next die then has only 1 chance to be right because MUST be the other number, so 1/6. You multiply those together, 1/18.

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.

   The fact that there are five kids doesn't matter. We don't care what the other three are. Since the chance of having a male is 1/2, and chance of the other being male is 1/2, the chance they both are is 1/2 * 1/2 = 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.

   Since it is six times, and you MUST get the pattern H H H T T T, then you just must multiple the probabilities together. Getting three heads is 1/2 * 1/2 * 1/2, and the same for getting three tails. So (1/2)^6 = 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.

   This one is tricky and my logic really fails.. but I have ideas as this one is simple enough. We have four ways this can look

   • HHHTTT
   • THHHTT
   • TTHHHT
   • TTTHHH

   ... I don't see how this goes to 5/16ths... sorry. I get 4/64 = 1/16

5. Hmm... But the P of getting two 3's is 1/36...How's that different than #1?

   You must get both to be 3, which means each die has only 1 right answer out of 6 possible. In the first one, one die has two choices to be right, and the second one has one chance to be right dependent on what the first one rolled.

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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ratbastid

Oh, I see the difference between this and the earlier question now, thanks.

I agree; I believe as stated, the answer is 1/16.

Hain

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:

1. H H H H H T T T T T
2. H H T H T T H T T H

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.

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Hain

Well how did you do on the assignment?

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ratbastid

Yes--only because there are more possible outcomes that appear random than that appear ordered. Which tells you something about the nature of life, right there...