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probability

by pemdas » Wed Nov 30, 2011 2:42 pm
A group of ten people consists of six men and four women. If six people are selected, what is the probability that at least two of them are women?

A. 35/43
B. 37/42
C. 33/44
D. 35/42
E. 32/43
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by kanwar86 » Wed Nov 30, 2011 3:08 pm
IMO, correct answer is B)
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by pemdas » Wed Nov 30, 2011 3:09 pm
please explain
kanwar86 wrote:IMO, correct answer is B)
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by vishal.pathak » Wed Nov 30, 2011 3:11 pm
pemdas wrote:A group of ten people consists of six men and four women. If six people are selected, what is the probability that at least two of them are women?

A. 35/43
B. 37/42
C. 33/44
D. 35/42
E. 32/43
Reqd prob = 1 - P(at most 1 women)
P(at most 1 women) = P(no women) + P(1 women)
= ( 6C6 + 6C5*4C1 ) / 10C6
= ( 1 + 24 )/ 10*3*7
= 5/42
So Reqd prob = 37/42 IMO B
Last edited by vishal.pathak on Wed Nov 30, 2011 3:17 pm, edited 1 time in total.
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by kanwar86 » Wed Nov 30, 2011 3:14 pm
pemdas wrote:A group of ten people consists of six men and four women. If six people are selected, what is the probability that at least two of them are women?

A. 35/43
B. 37/42
C. 33/44
D. 35/42
E. 32/43
Approach: You need to select at least two women
Consider three cases: (4M, 2W); (3M, 3W); (2M, 4W)
Favorable set of events will be the summation of above three case
that is, 6C4.4C2+ 6C3.4C3 + 6C2.4C4.............1)
Total no. of events will be = 10C6..............2)
Reqd. prob = 1)/2)

Hope that helped!
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by pemdas » Wed Nov 30, 2011 3:20 pm
great, i've made this up for myself to test concepts on prob/comb

and it's B

my take is like (useful if the q. is modified to "at least three women" or other cases)
ways to select 2 women out of 4 and 4 men out of 6 -> 4C2*6C4=6*15
ways to select 3 women out of 4 and 3 men out of 6 -> 4C3*6C3=4*20
ways to select 4 women out of 4 and 2 men out of 6 -> 4C4*6C2=1*15
total=185 ways
____________
ways to select 4 people out of 10 -> 10C6=210

Required probability 185/210=37/42
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by LalaB » Mon Dec 05, 2011 11:18 am
pemdas wrote:A group of ten people consists of six men and four women. If six people are selected, what is the probability that at least two of them are women?

A. 35/43
B. 37/42
C. 33/44
D. 35/42
E. 32/43
Pemdas, lets go farther :)

Now imagine that this group of ten people (six men and four women) wants to sit in a circular table, and no 2 women want to seat together. in how many different ways can u seat them down? :)

and btw, after they sit down in a table, they will vote for a senior board of 4 ppl of equal number of men and women. how many ways are possible?
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by pemdas » Mon Dec 05, 2011 11:26 am
you need to post this separately for others to contribute as well
LalaB wrote:
pemdas wrote:A group of ten people consists of six men and four women. If six people are selected, what is the probability that at least two of them are women?

A. 35/43
B. 37/42
C. 33/44
D. 35/42
E. 32/43
Pemdas, lets go farther :)

Now imagine that this group of ten people (six men and four women) wants to sit in a circular table, and no 2 women want to seat together. in how many different ways can u seat them down? :)

and btw, after they sit down in a table, they will vote for a senior board of 4 ppl of equal number of men and women. how many ways are possible?
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