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A committee of 3 has to be formed randomly from a group of 6 [#permalink]
 20 Mar 2012, 16:19
Question Stats:
46% (02:43) correct
53% (00:55) wrong based on 1 sessions
A committee of 3 has to be formed randomly from a group of 6 people. If Tom and Mary are in this group of 6, what is the probability that Tom will be selected into the committee but Mary will not? A. \frac{1}{10}B. \frac{1}{5}C. \frac{3}{10}D. \frac{2}{5}E. \frac{1}{2} Look for any of the three variants: Tom-notMary-notMary, notMary-Tom-notMary, notMary-notMary-Tom. The probability of one such variant is \frac{1}{6} * \frac{4}{5} * \frac{3}{4} = \frac{1}{10} . The answer to the question is 3*\frac{1}{10} = \frac{3}{10} . Why can't this be 5C3/ 6C3 = 1/2? Where did I make mistake when I applied combination formula approach? I know that this leads to wrong answer.
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Here is why 5C3 is wrong: 5C3 is the total number of ways to select 3 people from 5. If Tom is one of the 5 people, 5C3 does not count ONLY those ways in which Tom will be included. It counts the number of ways in which 3 people can be selected from 5 people (including Tom). This will include selections which do not include Tom, and so overstate the number of ways of always including Tom in the selection. The correct solution: Tom will always be selected, so select the other 2 people from the remaining 4 people. Number of ways = 1 * 4C2 = 6 Total number of ways to select 3 people from 6 = 6C3 = 20 Therefore probability = 6/20 = 3/10
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teal wrote: A committee of 3 has to be formed randomly from a group of 6 people. If Tom and Mary are in this group of 6, what is the probability that Tom will be selected into the committee but Mary will not?
A. \frac{1}{10}
B. \frac{1}{5}
C. \frac{3}{10}
D. \frac{2}{5}
E. \frac{1}{2}
Look for any of the three variants: Tom-notMary-notMary, notMary-Tom-notMary, notMary-notMary-Tom. The probability of one such variant is \frac{1}{6} * \frac{4}{5} * \frac{3}{4} = \frac{1}{10} . The answer to the question is 3*\frac{1}{10} = \frac{3}{10} .
Why can't this be 5C3/ 6C3 = 1/2? Where did I make mistake when I applied combination formula approach? I know that this leads to wrong answer. You can solve this problem with direct probability approach as well. We need the probability of T, NM, NM (Tom, not Mary, not Mary). P(T, NM, NM)=3*\frac{1}{6} * \frac{4}{5} * \frac{3}{4}=\frac{3}{10}, we are multiplying by 3 since {T, NM, NM} scenario can occur in 3 different ways: {T, NM, NM}, {NM, T, NM}, {NM, NM, T}. Answer: C.
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Manager
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Nice problem.....
Went for E first...... but correct answer is C.....
@Bunnel: respect man!
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A committee of 3 has to be formed randomly from a group of 6 [#permalink]
07 Jun 2012, 23:45
another way
T , M , A , B, C, D be the six members where T = Tom and M= Mary
First spot : let first person selected be tom so = 1/6 = Tom is selected from 6 people so M, A, B, C, D remain
Second spot : 4/5= mary cannot be selected so only 4 people can be selected from 5 ,after that only 4 people remain
Third spot : 3/4 = Again Mary cannot be selected ,
now it is not necessary that T will be picked first , we can have _ T _ or _ _ T or T _ _ , so we have to make order of the same elements irrelevant . The non mary candidates are same so we can have only 3 cases i.e [ Non mary , Non mary , T ] and [non mary , T , Non mary] and [ T , Non mary , Non mary ] .so 3!/ 2! , which equals 3
Just like arranging the letters of the word [ ALL ] = ALL ,LAL, LLA . only 3 cases possible , which is 3!/ 2! ( divide by the number of same elements)
there fore : 1/6 * 4/5 *3/4 * 3!/2! = 3/10
Its good to know all methods in case you need them, cheers !!
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A committee of 3 has to be formed randomly from a group of 6 [#permalink]
06 Jan 2013, 13:09
Mary is not to be selected, and Tom is to be selected.
Thus tom and mary are out of picture in selection (tom selected, marry rejected)
Problem boils to selecting other 2 members out of remaining 4 which can be done in 4C2 ways.
Answer 4C2/6C3 = 3/10
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A committee of 3 has to be formed randomly from a group of 6
[#permalink]
06 Jan 2013, 13:09
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