Probability question...

[dentwannabe]

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Maybe some of the QR-inclined folks could help me with this...

What is the probability of six tails out of nine tosses of a fair coin?

topscore's explanation is
(1/64)*(1/8)*84 = 21/128

by the way, they got 84 from nCr=(9*8*7*6*5*4)/(6*5*4*3*2*1).

OK... I thought it would just be 1/64 times 1/8, the probability of getting the tails times the prob. of not getting tails... But they multiply by 84... Where do they get that whole nCr thing from?? I dont see it in my Kaplan white book.

Thanks

 

[dentwannabe]

also..

does anyone know of any good book for the QR section.. kaplans is pissing me off.. theres no probability, no interest, profit problems... 😡

 

[naithgirw]

Maybe some of the QR-inclined folks could help me with this...

What is the probability of six tails out of nine tosses of a fair coin?

topscore's explanation is
(1/64)*(1/8)*84 = 21/128

by the way, they got 84 from nCr=(9*8*7*6*5*4)/(6*5*4*3*2*1).

OK... I thought it would just be 1/64 times 1/8, the probability of getting the tails times the prob. of not getting tails... But they multiply by 84... Where do they get that whole nCr thing from?? I dont see it in my Kaplan white book.

Thanks

Does this help...

Six tails - Three heads

Therefore P = C (9,6) times C (3,3) times (1/2)raised to the sixth, times, (1/2) raised to the third

Hope this helps...if not, I'll try to break it down further

 

[naithgirw]

Check out this Website:

http://www.stetson.edu/~jrasp/280sol07.htm

 

[cdpiano27]

I will explain this question as clearly as possible.

We are trying to find Pr(6 tails out of 9 tosses of a fair coin).

First of all, this is a binomial distribution question.

This is because:
For each toss: we are considering tails to be our success let us call this p. and we are considering heads to be our failure: let us call this q which is basically (1-p): remember the sum of probaiblities must always be equal to 1.

Now we have nine tosses: - - - - - - - - -.
Our 6 successes could occur in many ways. in slots 1-6 slots 1, 2, 3, 4, 5, 7: 1, 2, 3, 4, 5, 8, etc. In fact since order is not important we have n total slots (in this case 9) and we are choosing for these successes (tails) to occr in ANY 6 places where order DOES NOT MATTER This is how we get 9 choose 6 (9 nCr 6). Now Our successes probability p which is Pr(tails) = 1/2 since the coin is FAIR. Our failure probaility q is (1 - 1/2) = 1/2. Now our success probability occurs x = 6 times so we have (1/2)^6 = 1/64 and our failure probability occurs (n-x) or (9-6 ) = 3 times so we have (1/2)^3 = 1/8. Putting this together is how we get (9 nCr 6) * (1/64) * (1/8).
Now recall that 9 nCr 6 is 9! / 6! 3! (Note: basic format is n! / [x! (n-x)!]) so ( 9 X 8 X 7 X 6 X 5 X 4 X 3 X 2 x 1 ) / ((6 x 5 x 4 x 3 x 2 x 1 ) x (3 x 2 x 1)) = (9 x 8 x 7) / (3 x 2 x 1) = 504 / 6 = 84.
Our answer is 84 / 512 = 21 / 128. QED

We can look at this is having 9 independent Bernoulli trials with success probability 1/2.

When we have n Bernoulli trials where in this case n = 9, we are ALWAYS dealing with a BINOMIAL DISTRIBUTION which has probability density function (pdf) ( n nCr x) * (p^x) * (1-p)^(n-x).
Where n is number of total trials, x is number of successes, and n-x is number of failures.

Now if we did not have a fair coin, then the problem would specify that our probaibility of getting a tail is let us say (2/3). here success probability p is (2/3) and failure probaiblity q Pr(heads) is (1/3).

So in this case our pdf would be (9 nCr 6) * (2/3)^6 * (1/3)^3

I do not know what math is on the DAT but if there is a lot of probability study the basic distributions, Binomial, geometric, Poisson, Bernoulli, negative binomial and when to use them according to the problem.
I doubt that anything calculus based shows up since it is more of a science test. Anyway the hardest section is definitely PAT!

But this is a pretty straightforward, basic question.

Hope this helps.

 

[dentwannabe]

niath, cdpiano.. thanks a lot. i checked out the link and good clear explanation cd...

it makes sense now.

thanks.

 

[Babi3Ducki3]

I'm also having trouble w/ probability problems. I can't tell when it's a permutation or combination or whatever else. Can anyone help me?

 

[naithgirw]

I'm also having trouble w/ probability problems. I can't tell when it's a permutation or combination or whatever else. Can anyone help me?

Hope this helps: http://www.gomath.com/algebra/probability.php

 

[r0entgen]

i just wanted to let you know...

the topscore math section is really not an accurate portrayal of the real dat math...

i've always though math to be very easy...and then i took the topscore and i kept getting low 20s...

then comes the real dat and i got a 28 on the math...

the real dat math is very straightforward, and NOTHING like what topscore demands of you...

i took the topscore tests..and they required all these odd formulas that we never use...but the real dat just requires the basics...

 

[Muzikman712]

Learned a simple way to do this problem in Genetics.....Here goes,

Use Binomial Expansion to expand (a+b)^9 (to the power of nine since you toss the coin nine times)

-Let a=probability of landing tails = 1/2
-Let b- probability of landing heads = 1/2

-When you expand this you get: a^9 + 9a^8b + 36a^7b^2 + 84a^6b^3 + 126a^5b^4+ 126a^4b^5 + 84a^3b^6 + 36a^2b^7 + 9ab^8 + b^9

-There's a simple trick to find the coefficients and exponents (to find coefficients: 1. The coefficient of the 1st term is always 1.
2. To determine the coefficient of the next term in the binomial expansion a. Multiply the coefficient of the preceeding a by the exponent a in that term. b. divide that by the number indicating the position of the term in the series. The first coefficient is one, multiplied by its exponent 9 gives you 9. Divide this by one since its the 1st term in the series and you get a coefficient of 9 for the second term, etc...(for the third term you do 9 times 8 divided by 2 since its the second term in the series giving you 36)

-Because you want the probability of getting 6 tails choose the term where a (from let statement) is to the power of 6 (84a^6b^3). Plug in the probabilities for a and b and you get 84(1/2)^6(1/2)^3 = .164

-This is soooooo easy to do once you do a couple and its a great way to find the probabilities of maaaaany different types of mendelian genetics crosses.

-Hope that helps you

😉

 

[luder98]

Here is the link to the old thread

 

[iminnow]

also..

does anyone know of any good book for the QR section.. kaplans is pissing me off.. theres no probability, no interest, profit problems... 😡

Try "Math for Standardized Exams" by Cliff Notes. I think Amazon.com has it for 9 bucks. Also try some GMAT math or entry college SAT type math with story problems. Don't forget basic Trig. 🙂

 

[cdpiano27]

Of course that method is totally valid!

But here is why it works!

First of all the binomial expansion is defined as follows:

Let p = success probability
q = failure probability = (1-p)

It is known by binomial expansion that in general:

(q + p)^n = (n nCr 0) p^0 * (1-p)^(n-0) + (n nCr 1) * p^1 * (1-p)^(n-1) + (n nCr 2) * p^2 * (1-p)^ (n-2) + .......... + (n nCr n-1) * p^(n-1) * (1-p)^(1) + (n nCr n) * p^n * (1-p)^0.

Note that that the first term (n nCr 0) * p^0 * (1-p) ^ (n-0) = p(0) and (n nCr 1) * p^1 * (1-p)^(n-1) = p(1). Hence in general p(x) = (n nCr x) * p^x * (1-p)^(n-x). So now applying this to the specific problem, we wanted the Pr(6 tails in 9 tosses) where p is our success, that is, Pr(tail) and q is our failure, that is, Pr(head) which is 1-p.

So we need p(6) which is simply (9 nCr 6) * (.5) ^6 * (1-.5)^3.

Now notice to other important facts, which I do not think would show up on the DAT but are fundamental:

the addition of all probabilities from a pdf must sum to 1 (the sample space).

so summation over x p(x) = summation from x = 0 to n (n nCr x) X p^x * (1-p)^(n-x) = (q + p)^n = 1^n = 1. Which will ALWAYS hold true.

To see this application more clearly,
Finally, for a more difficult problem from Casella and Berger's Statistical Inference Chapter 1 (taught in a masters level statistical theory / probability course) which will NEVER show up in the DAT (these are proofs) but is useful to see how this theorem can be "tweaked" to get what you want:

Exercise
1.27 Verify the following identities for n>=2.

(a) summation k = 0 to n (-1) ^ k * (n nCr k) = 0.

Proof: Notice that this summation is EXACTLY in the binomial expansion format. That is first look at the fact that (a + b) ^n = summation over k=0 to n (n nCr k) * a^k * (b)^ (n-k). Now we notice in this problem that a is
-1 and although there appears to be nothing for b it is 1^(n-k) = 1 since 1 raised to any power is always 1!. and we clearly have n nCr k. so this summation can be written as (1 + -1)^n = 0^n = 0. And this identity is verified. QED

(b) summation x = 1 to n x * (n nCr x) = n * 2 ^(n-1)

Proof: The trick is to notice that by binomial theorem

( x + 1)^n = (n nCr 0) * (x^0) * (1^n) * (n nCr 1) * (x^1) * (1^(n-1)) * (n nCr 2) * (x^2) * (1^(n-2)) +......+ (n nCr n) * (x^n) * (1^0)

= (n Ncr 0) + (n nCr 1) * x + (n nCr 2) * x^2 +........ + (n nCr n) * x^n:

Now if you are clever you will notice that d / dx (x +1)^n = n(x+1)^(n-1).
So what will this do? Why did we start out with (x+1)^n? Notice if we let x = 1 after taking the derivative of (x +1)^n with respect to x we have n * 2^(n-1) which is what we are trying to get!

So taking the derivative with respect to x of the above binomial expansion gives:

(n nCr 1) + (n nCr 2) * 2x + (n nCr 3) * 3x^2 + (n nCr 4) * 4x^3 + .....+ (n nCr n) * nx^n-1.

letting x = 1 gives us
(n nCr 1) + (n nCr 2 ) * 2 + (n nCr 3) * 3 + (n nCr 4) * 4 +.......(n nCr n) *n = summation x = 1 to n x * (n nCr x).

So we have shown the equality. QED

(c) Show that summation x = 1 to n (-1)^(x+1) * x * (n nCr x) = 0

Proof: This one is very easy once you know the trick from part b. Notice that the left hand side is simply the expansion in b with the even terms (2nd, 4th, etc) as negative. In general we know that from part b that d / dx (1 + x) ^n = n * (1 + x)^(n-1) = (n nCr 1) + (n nCr 2) 2x + (n nCr 3) 3x^2 + (n nCr 4) * 4x^3 + ..... + (n nCr n) * nx^n-1.

To make the 2nd, 4th, and even terms negative in part b simply let x = -1. So this expansion is equivalent to n * ( 1+ -1)^(n-1) = n * 0^(n-1) which is obviously 0.

QED


I doubt you will get a question like this on the DAT, but I believe that it is very valuable to see where the underlying theory comes from especially when you can see calculus tricks and how the binomial theorem comes to play. This will just make the word problems MUCH easier!

 

[ddsnp]

👍 woah. i'm impressed with your math skills.

I know of this great book by kaplan. I don't know why they don't have a good description in our DAT book. THe book is GRE & GMAT exams math workbook and it is 18 dollars. Great book!

 

[cowsgomoo]

I hope there is not a lot of this probability junk on the real thing. I knew I should have paid attention in statistics class.