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Q: Probability Theory ( Answered ,   1 Comment )
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 Subject: Probability Theory Category: Science > Math Asked by: asadmu-ga List Price: \$10.00 Posted: 17 Jan 2005 17:45 PST Expires: 16 Feb 2005 17:45 PST Question ID: 458945
 ```Suppose that you continue to toss a coin, having probability p of coming up heads, until atleast one head and one tail have been observed. (a) Find the expected number of tosses needed (b) Find the expected number of tosses that come up heads Please show all work (even the small steps)```
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 Subject: Re: Probability Theory Answered By: elmarto-ga on 17 Jan 2005 19:45 PST Rated:
 ```Hi asadmu! In order to solve this problem, we'll make use of the geometric distribution. Geometric Distribution http://arnoldkling.com/apstats/geometric.html The geometric distribution answers a question that is similar to yours: given an experiment with two possible outcomes (success-failure, or heads-tails), what's the probability that the experiment must be repeated "x" times before a success is observed? The geometric distribution is determined by the following equation: f(x) = p*(1-p)^(x-1) , for x>=1 For example, what's the probability that only 1 toss is needed? It's f(1) = p*(1-p)^0 = p; or in other words, the probability of observing a success. Similarly, the probability of needing 2 tosses turns out to be p*(1-p), which is the probability of observing one failure [probability (1-p)] and one success [probability p]. As you can see in the link I provided above, the expected value of a geometrically distributed random variable is 1/p. Let's see how this applies to your question. Let's define p as the probability of getting heads, so (1-p) is the proabbility of getting tails. It's clear that you will need to make at least 2 coin tosses in order to see at least one heads and one tails. Let's assume now that we have already made the first toss and it turned out to be tails. We're now about to make our second toss. Since we've already observed a tails, we should now toss the coin until a heads is observed. Therefore, notice that, given that the first toss turned out to be tails, the number of tosses we now need in order to see a heads is simply a random variable with a geometric distribution, where "success" is getting a heads, which has probability p. Since we know the expected value of a geometrically distributed random variable, we can now say that if the first toss is tails, we'll need 1/p more tosses in order to get a heads. So we'll need a total of (1/p + 1) tosses in this case. What happens, on the other hand, if the first toss results in a heads? Now we should continue tossing the coin until a tails is observed. Therefore, the number of tosses we need to make now is a random variable with a geomtric distribution, where "success" is getting a *tails*, which has probability (1-p). Therefore, the expected value of the number of tosses we'll need to make (counting from the second one) is simply 1/(1-p). So we'll need a total of (1/(1-p) + 1) tosses in this case. Now we can easily calculate the expected number of tosses needed to get both a heads and a tails, before the first toss is made. We already know that in the first case, we'll need 1/p+1 tosses, and in the second case, we'll need 1/(1-p)+1. Since the probability of the first case is (1-p) (the first case is observing a tails in the first toss) and the probability of the second case is p, the expected number of tosses needed top observe both heads and tails is: (1-p)*(1/p+1) + p*(1/(1-p)+1) = (1-p)/p + (1-p) + p/(1-p) + p = (1-p)/p + p/(1-p) + p + 1 - p = (1-p)/p + p/(1-p) + 1 So the answer to question (a) is (1-p)/p + p/(1-p) + 1 We can use a similar framework in order to answer question (b). Let's assume as before that we've already made the first toss and are about to make the second one. The first toss results in a tails. How many heads should we expect to see by the time the experiment ends? It's clear that if the first toss resulted in a tails, then we will only see 1 heads, because when we get heads the experiment will end (we already got a heads and a tails). What happens now if the first toss comes up heads? We saw before that the expected value of tosses needed until the coin comes up tails in this case is 1/(1-p). Or, in other words, we expect the coin to come up heads 1/(1-p)-1 times (after the first toss) before we get a tails. Therefore, counting the first toss, in this case the expected number of heads is 1/(1-p)-1+1=1/(1-p). Finally, in order to get the expected number of time that coin will come up heads before making the first toss, we simply multiply the expected value in each case by the probability of each case and sum them, just like before: (1-p)*1 + p*(1/(1-p)) =p/(1-p) + (1-p) So the expected number of tosses that come up heads is p/(1-p)+(1-p). Google search terms "geometric distribution" ://www.google.com/search?hl=en&q=%22geometric+distribution%22 I hope this helps! If you have any doubt regarding my answer, please don't hesitate to request clarification before rating it. Otherwise, I await your rating and final comments. Best wishes! elmarto``` Request for Answer Clarification by asadmu-ga on 17 Jan 2005 23:15 PST ```Hello almarto thanks for the reply.. i need some clarififcation: This is the way i though about the question: for part a) Let N = # of tosses needed N>=2 Pr{N=k)=Pr{N=k|head first}*Pr{head first}+Pr{N=k|tail first}*Pr{tail first} =(p^(k-1))*(1-p)+((1-p)^(k-1))*p then E[N]= sum from k=2 to infinity k*Pr{N=k) and then i get (1/p(1-p))-1 Which is different from yours.. Your logic is the same as mine but we get different answers.. Can you please scrutinize this? Did i go wrong in the summation? part b) M=# of tosses that come up heads, then from the result in (a) M>=1 Pr{M=k}=Pr{N=k+1|head first}Pr{head first}=p^k(1-p) then Pr{M=1}=Pr{N=2|head first}*Pr{head first} + Pr{N=2 or 3 or 4 ....|tail first)*Pr{tail first} =p(1-p)+[(1-p)+(1-p)^2+....]*p=p(1-p)+1-p=1-p^2 so E[M]=1*Pr{M=1} + summation of k=2 to infinity k*Pr{M=k) =1-p^2+summation of k=2 to infinity kp^k(1-P)=1-p+(p/1-p) Please chack out my answer... Please help! Thanks Asad``` Clarification of Answer by elmarto-ga on 18 Jan 2005 07:40 PST ```Hi asadmu! As a matter of fact, we got the exact same results :-) My answer for (a) was: (1-p)/p + p/(1-p) + 1 Summing: (1-p)^2 + p^2 -------------- + 1 p(1-p) 1-2p+p^2+p^2 = -------------- + 1 p(1-p) 1 - 2p(1-p) = -------------- + 1 p(1-p) 1 = --------- - 1 p(1-p) which is the results you reported. In question (b) I got: p/(1-p) + (1-p) while you state you got 1-p+(p/1-p), which is of course the same. It's true that I used a more "intuitive" approach to explain my answer, but we did basically the same steps in order to arrive to the answer. If you have any further questions, please let me know. Best wishes! elmarto```
 asadmu-ga rated this answer: `Thanks that was great...!`

 Comments
 Subject: Re: Probability Theory From: readerlife-ga on 08 Dec 2005 23:56 PST
 ```I've always been confused by this question. You take your coin and flip it, having a 1/2 chance of it landing on heads. You flip it again, having a 1/2 chance of it landing on heads. So on and so forth until your 100th flip. Now assume you have managed to get 99 heads up to this point. What is the probability to get another head in the 100th toss? Theoretically, because each toss is independent, it should be again 50%. But probability theory also tells us that in the long run, the tendency should be half heads and half tails. So, we got 99 heads already, but it doesn't increase the chance of getting a tail next toss. Then, how could the probability structure of half-half be demonstrated anyway?```
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