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PS : Probability # 3

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CEO
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Joined: 15 Aug 2003
Posts: 3470
Followers: 60

Kudos [?]: 665 [0], given: 781

PS : Probability # 3 [#permalink] New post 14 Sep 2003, 02:22
00:00
A
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C
D
E

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(N/A)

Question Stats:

0% (00:00) correct 0% (00:00) wrong based on 0 sessions
my answers are in yellow..check 'em later..

y'all have a total of 1.5 minutes to do both of these.

Time starts now !

3 men, 3 women, select 4 people, what is the probability that the number of men is equal to the number of women?

Is it 3/5 ?

Five alphabets D I G I T can be made into how many combination so
that the two I will be separated at least by one alphabet?


Do we consider each I to be unique?

5! - (the # of ways in which the two "unique" I's are together)

5! - 4! * 2 = 72

Is this correct ?


Explain your method please

thanks
praetorian
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Joined: 21 Aug 2003
Posts: 258
Location: Bangalore
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 [#permalink] New post 14 Sep 2003, 02:50
i took 2.5 minturtes

1) 3/5 is correct
No of way for two men and two women to be equal i.e. 2M,2W =
3C2*3C2 -----------1
Total way to select 4 people = 6C4 -------------- 2
Thus, Probability = eq 1/eq 2 = 3/5

2) Why do u consider two I's unique, they are same..!!
Total ways = 5!/2 = 60
Ways in which I are together = 4! =24
Total ways in which I are not together = 60 -24 = 36
The same question can be made more interesting (or intimidating.. :evil: :evil: ) if its phrased like this:
Q - If five alphabets of DIGIT are used to form all possible words and one such word is picked at random, what is the probability that two I's are together in it. (Ans: 36/60 = 3/5)
-Vicks
ps: praet how do u solve that PS: oranges problem.. it is making me mad... give ur solution..
  [#permalink] 14 Sep 2003, 02:50
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PS : Probability # 3

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