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 July 28th, 2013, 12:55 AM #1 Member   Joined: Sep 2011 Posts: 99 Thanks: 1 Probability doubts I have encountered some problem with the probability qns. I have gotten 5/20 x 4/19= 1/19 for b (i) and (1/4 x 15/19) + (1/4 x 4/19) = 1/4 for b (ii). The answer is with the multiplication of 2 to 5/20 x 4/19 and to (1/4 x 15/19) + (1/4 x 4/19) as stated below. I do not understand why 2 must be multiplied to the probability? I need help in the explanation. Your help will be greatly appreciated. Many thanks. Here is the qns: Sam bought a set of 20 past year test papers. The set contains 4 subjects; English, Mathematics, Physics and Chemistry. There are 5 test papers for each subject numbered 1,2,3,4 and 5. (a) A test paper is chosen at random. Find , as a fraction, the probability that it is a mathematics test paper and numbered 5 or an English test paper. (b) The test paper is placed back into the set and now, two test papers are chosen at random. Find as a fraction, in its lowest terms, the probability that (i) they are both chemistry test papers. (ii) at least one is a mathematics test paper. Answer to (a) 1/20 + 1/4 = 3/10 (b) (i) 5/20 x 4/19 x 2 = 2/19 (b) (ii) (1/4 x 15/19 x 2) + (1/4 x 4/19 x 2) = 1/2
 July 28th, 2013, 04:57 AM #2 Senior Member   Joined: Jun 2013 From: London, England Posts: 1,316 Thanks: 116 Re: Probability doubts The correct answers to b (i) and b(ii) are: (i) (5/20) * (4/19) = 1/19 (You are correct. There is no reason to multiply this by 2.) (ii) 1/4 + (3/4)*(5/19) = 17/38 The answers given are definitely wrong. But, your calculations for b (ii) are not quite right. Her'e's how to think about b (ii): First, there is 5/20 = 1/4 probability that the first paper is maths. The second paper can then be anything. Note that this includes the possibility that both are maths. But, if the first paper is not maths (15/20) = 3/4, then there is still a chance that the second is maths (5/19). So, the answer is the sum of these. (5/20) + (15/20)(5/19) = 17/38. Note that an alternative way to work this out is to calculate separately: prob the first is maths and the second isn't + prob first is not maths and the second is + prob first and second are both maths. This will give you the same answer of 17/38.
 July 28th, 2013, 05:22 AM #3 Senior Member   Joined: Jun 2013 From: London, England Posts: 1,316 Thanks: 116 Re: Probability doubts One of the things I like to do is think of different ways to do these probability questions and then, hopefully, they all give the same answer. So, here's another solution for b (ii): Calculate the probability that neither is maths. Then subtract this from 1 to get the probability that at least one is maths: Prob neither is maths = (3/4)(14/19) = 21/38. So prob at least one is maths = 1 - 21/38 = 17/38. That settles it! And, another solution for b (i); Calculate the probability that both papers are the same, then divide this by 4, as they are equally like to be any subject. The first paper can be anything, and the second is the same with probablity 4/19. This is the prob that both papers are the same subject. So, there is 1/19 of both chemistry, 1/19 both maths etc.
 July 28th, 2013, 08:36 AM #4 Member   Joined: Sep 2011 Posts: 99 Thanks: 1 Re: Probability doubts Thank you so much for the help! Here, I have another qns. I have gotten( 1/6*5/6*6/6) + (6/6*5/6*1/6) + (6/6*1/6*1/6) = 11/36 for a(ii). I am not sure if my answer is correct or not as i find that my answer seems logical from the approach used in the previous doubts stated above. Three unbiased dice are thrown. Find the probability that they (a) (i) all show different numbers (ii) at least two show the same numbers. Answer for a(i) is 6/6 * 5/6 * 4/6 = 5/9 a(ii) is ( 6/6*1/6*1/6) + (6/6*1/6*5/6) = 1/6
 July 28th, 2013, 08:51 AM #5 Senior Member   Joined: Jun 2013 From: London, England Posts: 1,316 Thanks: 116 Re: Probability doubts The quick way is to note that (i) and (ii) are complements. I.e. either all dice are different or at least two are the same. The answer give to (i) is correct: 5/9. So, the answer to (ii) is 1 - 5/9 = 4/9. It is, of course, possible to crunch the answer to (ii) by looking at all the options where at least two are the same, but this is much more complicated than calculating the complement of this and subtracting from one.

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