Probabilities and Expected Value in Seating ArrangementsDate: 05/12/98 at 08:01:18 From: Ian Peters Subject: Probability Dear Dr. Maths, Three couples are seated at random around a round table. Let M denote the number of husbands seated next to their wives. Find: (a) the probability that M = m for m = 0,1,2 and 3 (b) the expected value of M using part(a) I know that for 1 couple siting together, the probability would be 2/ 5: say the husband sat down, then there are only 5 seats left, and his wife can sit on either side of him, so it would be 1/5 + 1/5. But then as I work to 2 couples, I don't know how to state the situation, and I got nonsense answers (sometimes the probability is greater than 1, which is impossible). And also, what does it mean by the expected value of M using part (a)? Can you give me one example, say, to find the E(M) as m = 2, please? Thanks for your time, Doctor. Ian Date: 05/12/98 at 19:36:23 From: Doctor Anthony Subject: Re: Probability For part (a), with a little systematic placing, we have: P(0) = 32/120 P(1) = 48/120 P(2) = 24/120 P(3) = 16/120 ------------------ Total = 1 The denominator 120 is the total number of permutations 5! = 120. Working out above results is done in the following way. For example: with P(3), we consider (aA) (bB) (cC) where a = husband, A = wife, and so on. Keep (aA) in fixed position. Then we can swap A and a, b and B, c and C, giving 2*2*2 different arrangements; and finally we could swap the b pair with the c pair, giving a further factor 2. All together, there are: 2*2*2*2 = 16 arrangements with all three seated together. In part (b), we find the expected value of M using part (a): Expected value = 0*P(0) + 1*P(1) + 2*P(2) + 3*P(3) = 0 + 1*48/120 + 2*24/120 + 3*16/120 = (1/120)[48 + 48 + 48] = 3*48/120 = 1.2 -Doctor Anthony, The Math Forum Check out our web site! http://mathforum.org/dr.math/ Date: 05/12/98 at 21:28:53 From: ian peters Subject: Re: Probability Dear Dr. Anthony, I still don't understand that why P(2) is not same as P(3). If P(2) is correct, that should mean that there are only 2 seats and 1 couple left -- so doesn't that mean the third couple will sit together as well? Also, why is P(1) not 2/5? Can you show me how you work P(1) and P(2) out, please? Thanks, Doctor Anthony. Ian Date: 05/13/98 at 11:35:15 From: Doctor Anthony Subject: Re: Probability Two pairs could separate a third pair: (aA) b B (cC) You can choose which two pairs in 3 ways, and then the a pair can be swapped, the b pair can be swapped, and the c pair can be swapped, giving: 3*2*2*2 = 24 ways With 1 pair together, you could have: (aA) c b B C You could choose the pair to be together in 3 ways. Then the a pair can be swapped, the b pair can be swapped, the c pair can be swapped. Also, keeping the the a pair together, you could swap the b pair with the c pair. This would give: 3*2*2*2*2 = 48 ways We have already shown that the three pairs together can occur in 16 ways. You could go through another exercise with no pairs together, but the number of ways is clearly 120 - 16 - 24 - 48 = 32 ways. I have checked this result, but it is somewhat tedious, requiring 4 lots of situations each giving 8 arrangements, totalling 32: A A A A c C c B C b c B b B b a a c b c a C B a In these, we keep A fixed and move a through three possible positions. Then for each diagram, we can swap the b pair, swap the c pair, and swap the b pair and c pair, giving 2*2*2 = 8 ways of arranging the letters. -Doctor Anthony, The Math Forum Check out our web site! http://mathforum.org/dr.math/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994-2013 The Math Forum
http://mathforum.org/dr.math/