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12 Jun 2004, 00:52
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There is a group of six people; among them are Mr X and Mr Y. Three persons are chosen at random without repetitions. Find the probability of choosing:
1) Mr X, but not Mr Y.
2) Both Mr X and Mr Y.
3) Neither Mr X nor Mr Y.
4) Either Mr X or Mr Y, but not both.
5) Either Mr X or Mr Y or both.
1) Mr X, but not Mr Y.
2) Both Mr X and Mr Y.
3) Neither Mr X nor Mr Y.
4) Either Mr X or Mr Y, but not both.
5) Either Mr X or Mr Y or both.
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12 Jun 2004, 01:40
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stolyar
1. C[2,4]/C[3,6] = 36/120 = 6/20 = 3/10.
2. C[1,4]/C[3,6] = 4*6*6/(6*120) = 1/5.
3. C[3,4]/C[3,6] = 1/5.
4. 2*P(X, but not Y) = 6/10 = 3/5.
5. P(X or Y, but not both) + P(both) = 3/5 + 1/5 = 4/5.
13 Jun 2004, 13:23
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Agree
(1) 0.3
(2) 0.2
(3) 0.2
(4) 0.6
(5) 0.8
(1) 0.3
(2) 0.2
(3) 0.2
(4) 0.6
(5) 0.8
Archived Topic
Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
Where to now? Join ongoing discussions on thousands of quality questions in our Quantitative Questions Forum
Still interested in this question? Check out the "Best Topics" block above for a better discussion on this exact question, as well as several more related questions.
Thank you for understanding, and happy exploring!
