Probability in Flipping CoinsDate: 2/25/96 at 19:16:27 From: Dana Kaye Subject: Probability Question I am looking for some help with a couple of probability questions. On Friday we had a substitute teacher who explained probability to us and gave us some questions. Since we share textbooks at school we can't take the textbooks home, and my parents don't have a clue. I am not looking for you to do my homework for me (Though that would be nice), but I would like some instruction on how to go about solving these problems. Here are the problems: 12. Six pennies are flipped. What is the probability of getting a. two heads and four tails? b. four heads and two tails? c. at least three heads? d. at most two tails? 13. A bag contains one red marble and one white marble. Marbles are drawn and replaced in a bag before the next draw. a. What is the probability that a red marble will be drawn two times in a row? b. In ten draws, what is the probability that a red marble will be drawn at least five times. Thanks for your help! Sincerely, Alexia Kaye Date: 2/26/96 at 15:22:46 From: Doctor Syd Subject: Re: Probability Question Dear Alexia, Hello! I'm glad you wrote for help -- that is what we are here for! Probability can be pretty cool once you understand the basic concepts, so let's figure out how to approach your problems and see where we can go from there. Okay, so if you are flipping 6 coins, and you want to know the probability of a certain event, the first and most important thing you need to know is what the probability that the coin will land on heads is, and what the probability that the coin will land on tails is. This is information that should be given to you in the problem - otherwise you don't have enough information. Usually these problems deal with a "fair coin," that is a coin in which there is an equal probability that it will land on head or tail. Thus, in the fair coin, the probability that the coin lands on heads is 1/2 because on the average, one out of every two times the coin lands on heads. Likewise the probability that the coin will land on tails is also 1/2. It is important to note that in the real world there really isn't an actual coin that is "fair." Most coins have probabilities that are nearly equal to 1/2, but are more like, for instance, .4899 probability for heads and .5101 probability for tails. Let's tackle that first problem. Suppose you want to know the probability that when you flip your six coins you will get the sequence:HHHHTT where H=heads and T=tails. Well, the total number of sequences is going to be 2^6 (that's 2 to the sixth power). This may be something you have seen before; if not, feel free to write back and I'll explain more. So, the probability of getting this one sequence is going to be 1/2^6, right? Well, now let's use the same general strategy to solve your problem. How many sequences have 4 heads and 2 tails in them? Certainly the sequence above does, and there are others: HHTTHH is an example. So how can we count them? Well, it turns out that there are "6 choose 2" such sequences. I'm not sure how much combinatorics (combinatorics is a fancy word for the subject of math that has to do with counting things like the number of sequences of H's and T's such that there are 4 H's and 2 T's!) you have had, so if you don't understand some of what I write, please write back so I can clear it up. Basically, we can think of the place of the tails as a distinguishing feature of the sequences with 4 heads and 2 tails. In other words, once we know where the tails are, since we have just 4 other slots left, those slots must be filled by heads, and we are okay. So, in how many ways can we put 2 T's into 6 different slots? The combinatorial answer is "6 choose 2" = 6!/ (4!)(2!). Thus, the number of sequences with 4 heads and 2 tails is 6!/ (4!)(2!). Thus the probability that we'll get such a sequence is: [6!/ (4!)(2!)]/2^6 = 6!/(4!)(2!)(2^6). Phew! That was a long answer to a seemingly innocent and simple question, eh?! Your second question is quite similar to the first, so I leave that to you. The third and fourth questions can be figured out using methods similar to the first two problems, but by extending just a bit. Remember that it is sometimes easier to figure out the probability that a certain event will NOT happen, and then once you know that, you can use the fact the probability that an event will happen + the probability that an event won't happen = 1 to figure out the desired probability. For instance, in problem c, it might be easier to figure out the probability that you will not get at least 3 heads. If you don't have at least 3 heads, what does that mean? It means you must have 0, 1, or 2 heads, right? So, what is the probability that you have 0 heads? The probability that you have 1 head? The probability that you have 2 heads? (These are things you can solve using exactly the same method we used to solve your first two problems). Since the events are mutually exclusive (that is, if you have a sequence with exactly 1 head, then it is impossible for it to have exactly 2 heads. Whoa! What a thought! Sometimes that goofy math terminology is expressing the simplest of notions!), the probability that you have 0, 1, or 2 heads is the probability of having exactly 0 heads + probability of having exactly 1 head + the probability of having exactly 2 heads. So, from here you can probably get an answer. I'm going to leave the marble questions to you, for now. See if you can figure out what to do with them, having done the coin problems. If I've said anything that is unclear, or you have any more questions, feel free to write back. -Doctor Syd, The Math Forum |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994-2013 The Math Forum
http://mathforum.org/dr.math/