My Math Forum Probability: Completely Lost... Central Limit Theorem(?)

 September 7th, 2018, 05:24 AM #1 Newbie   Joined: Sep 2018 From: Australia Posts: 3 Thanks: 0 Probability: Completely Lost... Central Limit Theorem(?) Hi everyone, This is my first post here and I'm really sorry if this isn't done right. I have an exam coming up and I've been going through the questions we've had so far in class and this is... I have no idea how to begin with it, let alone proceed. When I went to the uni drop-in Numeracy Centre, they were closed (today, Friday), and it's coming up early Saturday my time. The exam is on Monday The question is this: If you toss two coins 120 times, what is the probability that fewer than 36 of those tosses would result in two tails? I've been trying for days now to get my head around it and I haven't been able to. Showing my work: This was the week we were learning the Central Limit Theorem, so I assumed that was part of the solution - the distribution of the number of heads/tails must be a normal distribution, right? (I'm not sure why it would be, but that's my own folly. I looked up 'distribution of coin flips' and yeah, that there sure looked like a normal curve) My line of thinking was like this: the probability is the area proportion under the curve before, after or between two spot(s), right? And the possible amounts of double-tails that fulfill the conditions are 0, 1, 2 ... 35. So it's the area under the curve between the values for 0 to 35. Get the z-scores, get the values, subtract, yeet. There's your probability. Get z-scores. Mean? Hm. Well, the most common value should be the most likely value, and the probability of a single double-coin throw being both tails is 1/4, so it's most likely that 25% of the tosses are double-tails, and a quarter of 120 is 30, so the mean is 30. Now the standard deviation, and that's... uh. A number. 3? It's probably not 3. How do you get the standdev? Even with the standard deviation, where do you go from there? Is it like that the x-axis is the number of double tails, so all I really need is the area under the curve up to 35? Great, that's in the z table! Just need that pesky z-score. But if it's not that, then I completely have no idea. Nothing. I tried plugging it into Wolfram Alpha, but it just reads it as 'coins 120 times' (ie 120 flips of one coin), and even the formula it gave for 'Probability of x heads' was not something I know how to interpret (what does the bracketed 120 above x mean? No line betwixt 'em, so not a fraction, tumorrav, I thought) If you can give me any help I'd really appreciate it, even if it's to tell me I've gone in completely the wrong direction. Thanks for your time.
 September 7th, 2018, 03:50 PM #2 Global Moderator   Joined: May 2007 Posts: 6,806 Thanks: 716 The probability of 2 tails is $\frac{1}{4}$, so the probability that after n tosses you have k tail pairs is $\displaystyle p_{k,n}=\binom{n}{k}\left(\small\frac{1}{4}\right) ^k\left(\small\frac{3}{4}\right)^{n-k}$. What you want is $\displaystyle \sum_{k=0}^{35}p_{k,120}$. Last edited by skipjack; September 7th, 2018 at 09:39 PM.
September 7th, 2018, 04:12 PM   #3
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Quote:
 Originally Posted by mathman The probability of 2 tails is $\frac{1}{4}$, so the probability that after n tosses you have k tail pairs is $\displaystyle p_{k,n}=\binom{n}{k}\left(\small\frac{1}{4}\right) ^k\left(\small\frac{3}{4}\right)^{n-k}$. What you want is $\displaystyle \sum_{k=0}^{35}p_{k,120}$.
Sort of... I think what he wants is the normal approximation to the binomial distribution that arises from the central limit theorem.

Last edited by skipjack; September 7th, 2018 at 09:40 PM.

September 7th, 2018, 05:34 PM   #4
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Thanks for the help, you guys especially on a weekend, it means a lot. I'd assume the binomial distrib is what my professor wants? So I'd be grateful for it but if else, the exam lets us bring a page of notes so I can bring your explanation with me, mathman.

Quote:
 Originally Posted by mathman The probability of 2 tails is $\frac{1}{4}$, so the probability that after n tosses you have k tail pairs is $\displaystyle p_{k,n}=\binom{n}{k}\left(\small\frac{1}{4}\right) ^k\left(\small\frac{3}{4}\right)^{n-k}$. What you want is $\displaystyle \sum_{k=0}^{35}p_{k,120}$.
I'm not quite sure how to interpret ${k,n}$ and $\displaystyle \sum_{k=0}^{35}p_{k,120}$. To save, you having to explain it; if you don't want to, is there a name for the n atop k in brackets so I can look it up? I get the sum notation, just not the comma here $p_{k,120}$

Big thanks again

Last edited by skipjack; September 7th, 2018 at 09:42 PM.

 September 7th, 2018, 09:43 PM #5 Senior Member     Joined: Sep 2015 From: USA Posts: 2,529 Thanks: 1389 via the CLT this binomial distribution can be closely approximated by a normal distribution with parameters $\mu = \dfrac n 4 = 30$ $\sigma = \sqrt{n \dfrac 1 4 \dfrac 3 4} =\dfrac{3\sqrt{10}}{2}$ so the random variable $Z = \dfrac{X-30}{\frac{3\sqrt{10}}{2}}$ is a standard Normal, and the usual tables can be used to compute probabilities. Plugging in $X=36$ we find $P[\text{fewer than 36 TT rolls}] \approx 0.897$ Because the normal distribution is continuous and the binomial distribution is discrete a "continuity correction" is usually applied when using this approximation. You can google and read up on that.
 September 7th, 2018, 09:55 PM #6 Global Moderator   Joined: Dec 2006 Posts: 20,921 Thanks: 2203 The $\displaystyle \binom{n}{k}$ is a notation for the binomial coefficient.
 September 8th, 2018, 05:50 PM #7 Global Moderator   Joined: May 2007 Posts: 6,806 Thanks: 716 $p_{k,n}$ is a symbol I made up to represent k successes out of n tries.
 September 9th, 2018, 11:36 PM #8 Newbie   Joined: Sep 2018 From: Australia Posts: 3 Thanks: 0 mathman, romsek, skipjack, thank you so much for the help! Your explanations aided me immensely, thank you for being so thorough and clear. I owe a lot to all of you, including now at least 20% of my mark.

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