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Q: Statistics homework ( Answered ,   1 Comment )
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 Subject: Statistics homework Category: Reference, Education and News > Homework Help Asked by: as3411-ga List Price: \$7.50 Posted: 03 Nov 2002 12:28 PST Expires: 03 Dec 2002 12:28 PST Question ID: 97472
 ```Please help answering these questions: 1.In a binomial experiment the probability of succes is 0.06. What is the probability of 2 successes in seven trials? Please show work...answers are a. .06 b. .0554 c. .28 d. .0036 2.. The probaility that Pete will catch fish on a particular day when he goes fishing is 0.8. Pete is going fishing 3 days next week. The probability that Pete will catch fish on exactly 1 day is ? possible answers are a. .104 b. .008 c. .096 d.8 \$3.00 for each correct answer with work shown.```
 ```Summary: (a) and (c) Thank you for the question. I hope that I can help out here. Before rating this question, if you have any questions for me, please use the "Request Clarification" button that you see in your web browser to request clarification . (Sometimes the "Request Clarification" button is a little hard to find, but if you look for it on the page you are reading this answer, you should be able to find it). First, let's talk about combinations. The "choose" function gives the number of ways of choosing a subset of a set of elements. That is the number of "combinations" of k from n elements. Specifically, choose(n,k) is the number of subsets of k items included in a larger set of n items. choose(n,k) is usually written like this: n ( ) k But that is hard to read on a computer screen, so I will use choose(n,k) . In Matlab, choose(n,k) is called nchoosek(n,k). I definitely recommend the Matlab software, http://www.mathworks.com , for exploring statistics and many other things. I used Matlab for the computations here, although actually any calculator, or even pencil and paper (do students still use that?) would work. For example, suppose n is 4 and k is 2. If the larger set has elements {1,2,3,4}, then the possible 2-element subsets are: {1,2} {1,3} {1,4} {2,3} {2,4} {3,4} There are therefore 6 2-element subsets of a 4-element set, so choose(4,2)=6. The formula for computing choose(n,k) is n!/(k!(n-k)!) where n!=1*2*3*...*n. Since 4!=4*3*2*1=24, and 2!=2, we see that choose(4,2)= 24/(2*2)=24/4=6, as we computed. 1. A binomial experiment is a little like flipping a coin, not necessarily a fair coin. If we flip a coin that has a .06 chance of reaching heads and a .94 chance of tails 7 times, what are the odds we will get 2 heads? Unfortunately the question is ambiguous. It is not clear whether the questioner wants the probability of exactly 2 successes occuring or of at least 2 heads occurring. If the former, the probability is just choose(7,2) * .06^2 * .94^5 . If we imagine the coins flipped in order, there are choose(7,2) ways in which we can get exactly two heads. (That is, we can get heads on say, 1st and second flip; or 1st and 3rd flip; or 2nd and 4th flip, and so on). For each such occurrence of heads, the probability it occurs is .06^2*.94^5 . [For instance, the probability we get heads on first two flips and tails on remaining 5 flips is this] Now, choose(7,2) = 7!/((5!*2!))=21 (you can verify this by just enumerating the subsets: {1,2}, {1,3}, and so on. You should get 21 subsets. Hence, the probability we want is 21*.06^2*.94^5 = .055 rounded to two decimal points it's .06, so the answer is a. So if the questioner meant "exactly two successes" the answer is a. What if the questioner meant "at least two successes"? Then the probability is the same as 1 - P(zero or one successes) Now, zero successes happens with P=.94^7. 1 success happens with P=choose(7,1)*.06*.94^6 , so the probability we want is 1- (.94^7 + 7*.06*.94^6)= .0618 . So this answer would be (a) also. Hence, no matter what the question was supposed to mean, the correct answer is (a). QUESTION 2 --------- Very similar to the above problem. Suppose he goes fishing M, T, W. (Monday, Tuesday, and Wednesday). He can catch a fish on one day by either catching a fish M and not T and not W; or T and not M and not W; or W and not M and not T. Each of the three possibilities has P=.8*.2*.2 = .032 . Since there are three mutually exclusive possibilities, the result is 3*.032 = .096 . That is choose(3,1)*.032 . Hence the answer is (c). SEARCH STRATEGY --------------- Well, I know all this stuff from years of professional experience, and I am even more familiar with it since I just recently had to answer another similar question about combinations here: https://answers.google.com/answers/main?cmd=threadview&id=95478 . But once you think of it the right way and do a lot of practice problems, it's just common sense.``` Clarification of Answer by rbnn-ga on 03 Nov 2002 14:05 PST ```The commenter is correct. I'd computed that the probability of *exactly* two successes would be 0.0555 , which is close enough to the requirement for (b), 0.0554 that indeed it is true that: under the interpretation that *exactly* two successes are required, the answer would be (b). It's even MORE confusing though than this! That's because the two options given are: a) 0.06 and b) 0.0554 Well, the trouble is that "0.06" IS 0.0554 rounded to two decimal places. There is no way to know whether 0.06 is just a "rounding" of 0.0554 - it depends on how much precision is known about the "0.06" as the probability of success. In other words, the probability of success only has one significant digit, arguably, so perhaps a) is just saying we should not give the answer to greater precision than the question? But why would the question in 1) be phrased WITHOUT the word "exact" which WAS used in the same context in 2) ? So, I guess there really is no way to know the "correct" answer here. However, based on my experience with these kinds of homework questions, I NOW think the intended answer is PROBABLY (b) for 1, because: i. a) is the same as P=0.06, so it could be a common mistake the book is trying to trap you with. ii. It is unlikely at this level that issues of precision would be introduced, and iii. It is quite possible that the word "exact" was just left off the question statement iv. The agreement is fairly close, and the "exact" interpretation is easier mathematically than the "at least interpretation". In conclusion, I am going with (b) now, agreeing with the commenter, for question 1. Of course the math and derivation remain unchanged.```
 as3411-ga rated this answer: and gave an additional tip of: \$7.00 `Great...I needed the process, thank you!`
 `wouldn't the answer to number one be (b) if it meant exactly two flips?`